Atomic and Nuclear Physics Practice Questions — JAMB UTME — Page 7

Question 121Question →

A particle of constant mass has a de Broglie wavelength of $\lambda$ . If the kinetic energy of the particle is increased to nine times its initial value, what is its new de Broglie wavelength?

\frac{\lambda}{9}

\frac{\lambda}{3}

3\lambda

9\lambda

Show answer & explanation

Answer: $\frac{\lambda}{3}$

Answer

The new de Broglie wavelength is

\frac{\lambda}{3}

.

The de Broglie wavelength of a particle is given by

\lambda = \frac{h}{\sqrt{2mE_k}}

, showing that wavelength is inversely proportional to the square root of its kinetic energy. When kinetic energy is increased by a factor of 9, the square root factor becomes

\sqrt{9} = 3

, reducing the new wavelength to one-third of its original value (

\frac{\lambda}{3}

).

Step-by-Step Solution

1

Relate de Broglie wavelength to momentum and kinetic energy.

\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mE_k}}

Since kinetic energy is

E_k = \frac{p^2}{2m}

, linear momentum is

p = \sqrt{2mE_k}

.

2

Set up the ratio for the new kinetic energy

E_k' = 9E_k

.

\lambda' = \frac{h}{\sqrt{2m(9E_k)}} = \frac{h}{3\sqrt{2mE_k}}

Substitute the new kinetic energy into the de Broglie wavelength equation.

3

Express the new wavelength in terms of the initial wavelength

\lambda

.

\lambda' = \frac{\lambda}{3}

Comparing

\lambda'

to

\lambda = \frac{h}{\sqrt{2mE_k}}

gives

\lambda' = \frac{\lambda}{3}

.

Key Concept

De Broglie Wavelength and Kinetic Energy Relationship

Question 122Question →

A radioactive parent nucleus of Thorium, $^{232}_{90}\text{Th}$ , undergoes a natural radioactive decay series by emitting a total of $6$ $\alpha$ -particles and $4$ $\beta^-$ -particles to form a stable daughter isotope of Lead ( $\text{Pb}$ ). Calculate the number of neutrons present in the nucleus of the resulting daughter isotope.

Show answer & explanation

Answer: 126

Answer

The resulting daughter nucleus contains 126 neutrons.

Emitting

6

\alpha

-particles reduces the mass number by

6 \times 4 = 24

units and the atomic number by

6 \times 2 = 12

units. Emitting

4

\beta^-

-particles leaves the mass number unchanged while increasing the atomic number by

4 \times 1 = 4

units. Consequently, the daughter nucleus has a mass number

A = 232 - 24 = 208

and an atomic number

Z = 90 - 12 + 4 = 82

. The number of neutrons is

N = A - Z = 208 - 82 = 126

.

Step-by-Step Solution

1

Calculate the mass number (

A

) of the daughter nucleus after all emissions

A = 232 - (6 \times 4) = 208

An alpha particle carries away 4 mass units (

^{4}_{2}\text{He}

), while a beta-minus particle carries 0 mass units (

^{0}_{-1}\text{e}

).

2

Calculate the atomic number (

Z

) of the daughter nucleus after all emissions

Z = 90 - (6 \times 2) + (4 \times 1) = 82

Each alpha decay reduces nuclear charge by 2, and each beta-minus decay increases nuclear charge by 1 due to neutron-to-proton conversion.

3

Compute the number of neutrons (

N

)

N = A - Z = 208 - 82 = 126

The number of neutrons in any nuclide is given by subtracting the atomic number (protons) from the mass number (nucleons).

Key Concept

Mass and Atomic Number Conservation in Radioactive Decay Chains

Estimated Time:2m 0s

Question 123Question →

Two deuterium nuclei (

{^{2}_{1}\text{H}}

) undergo nuclear fusion to form a helium-3 nucleus (

{^{3}_{2}\text{He}}

) and a neutron (

{^{1}_{0}\text{n}}

) according to the nuclear equation:

{^{2}_{1}\text{H}} + {^{2}_{1}\text{H}} \rightarrow {^{3}_{2}\text{He}} + {^{1}_{0}\text{n}}

Given the atomic mass values:
Mass of

{^{2}_{1}\text{H}} = 2.0141\text{ u}

Mass of

{^{3}_{2}\text{He}} = 3.0160\text{ u}

Mass of

{^{1}_{0}\text{n}} = 1.0087\text{ u}

Taking

1\text{ u} = 931.5\text{ MeV}

, what is the energy released in this fusion reaction?

1.63\text{ MeV}

3.26\text{ MeV}

5.23\text{ MeV}

6.52\text{ MeV}

Show answer & explanation

Answer: $3.26\text{ MeV}$

Answer

The energy released in the fusion reaction is

3.26\text{ MeV}

.

The total mass of the reactants is

2 \times 2.0141\text{ u} = 4.0282\text{ u}

, while the total mass of the products is

3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

. Subtracting product mass from reactant mass gives a mass defect

\Delta m = 0.0035\text{ u}

. Multiplying this mass defect by the energy equivalent factor of

931.5\text{ MeV/u}

yields

3.26\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants

Total reactant mass

= 2 \times 2.0141\text{ u} = 4.0282\text{ u}

Two deuterium nuclei are fusing on the left side of the reaction equation.

2

Calculate the total mass of the products

Total product mass

= 3.0160\text{ u} + 1.0087\text{ u} = 4.0247\text{ u}

The reaction produces one helium-3 nucleus and one neutron.

3

Determine the mass defect (\Delta m)

\Delta m = 4.0282\text{ u} - 4.0247\text{ u} = 0.0035\text{ u}

Mass defect is the difference between total reactant mass and total product mass.

4

Calculate the energy released in MeV

E = 0.0035\text{ u} \times 931.5\text{ MeV/u} = 3.26025\text{ MeV} \approx 3.26\text{ MeV}

Multiplying the mass defect in atomic mass units (u) by

931.5\text{ MeV/u}

gives the total energy released.

Key Concept

Mass Defect and Energy Release in Nuclear Fusion

Estimated Time:1m 30s

Question 124Question →

An electron of mass $9.1 \times 10^{-31}\text{ kg}$ is moving with a velocity of $2.0 \times 10^6\text{ m s}^{-1}$ . What is the de Broglie wavelength of the electron? (Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ )

3.64 \times 10^{-10}\text{ m}

1.21 \times 10^{-57}\text{ m}

2.75 \times 10^9\text{ m}

3.64 \times 10^{-40}\text{ m}

Show answer & explanation

Answer: $3.64 \times 10^{-10}\text{ m}$

Answer

The de Broglie wavelength of the electron is

3.64 \times 10^{-10}\text{ m}

.

According to de Broglie's wave-particle duality hypothesis, the wavelength

\lambda

associated with a moving particle of mass

m

and velocity

v

is given by

\lambda = \frac{h}{m v}

. Substituting

h = 6.63 \times 10^{-34}\text{ J s}

,

m = 9.1 \times 10^{-31}\text{ kg}

, and

v = 2.0 \times 10^6\text{ m s}^{-1}

yields

\lambda = \frac{6.63 \times 10^{-34}}{1.82 \times 10^{-24}} = 3.64 \times 10^{-10}\text{ m}

.

Step-by-Step Solution

1

Calculate the linear momentum (

p

) of the electron

p = m v = (9.1 \times 10^{-31}\text{ kg}) \times (2.0 \times 10^6\text{ m s}^{-1}) = 1.82 \times 10^{-24}\text{ kg m s}^{-1}

The de Broglie wavelength depends directly on the momentum of the moving particle.

2

Apply the de Broglie wavelength formula

\lambda = \frac{h}{p}

\lambda = \frac{6.63 \times 10^{-34}\text{ J s}}{1.82 \times 10^{-24}\text{ kg m s}^{-1}} = 3.64285... \times 10^{-10}\text{ m}

Substitute Planck's constant and the calculated momentum to find the matter wavelength.

3

Express the answer in standard scientific notation with appropriate significant figures

\lambda = 3.64 \times 10^{-10}\text{ m}

Matching standard exam accuracy and precision.

Key Concept

de Broglie Wavelength of Matter Waves

Question 125Question →

Match each type of natural radioactive emission with its corresponding physical property regarding electric field deflection and penetrating ability.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alpha particle pairs with deflection towards the negative plate and highest ionizing power; Beta-minus particle pairs with deflection towards the positive plate and moderate penetrating power; Gamma ray pairs with no deflection in electric fields and highest penetrating power.

Alpha particles carry positive charge (

+2e

), causing them to deflect toward the negative plate in an electric field while producing intense ionization. Beta-minus particles carry negative charge (

-e

), deflecting toward the positive plate. Gamma rays are neutral photons, passing completely undeflected while possessing the greatest penetrating power.

Step-by-Step Solution

1

Determine the charge and nature of each radioactive emission.

Alpha particles are doubly positively charged helium nuclei (

+2e

), Beta-minus particles are negatively charged electrons (

-e

), and Gamma rays are neutral electromagnetic photons (

0

).

Electric field deflection depends directly on the sign and magnitude of particle charge.

2

Analyze behavior in an electric field and relative penetrating power.

Positively charged alpha particles move toward the negative plate and ionize strongly due to high mass/charge; negatively charged beta particles move toward the positive plate; uncharged gamma rays pass straight through and penetrate deepest.

Opposite charges attract in electric fields, while mass and charge dictate the rate of energy loss (ionization vs penetration) in matter.

3

Pair each emission to its exact matching property.

Alpha matches highest ionizing power and negative plate deflection; Beta-minus matches positive plate deflection; Gamma ray matches highest penetrating power and no deflection.

This correctly aligns each particle/wave type with its distinct physical behavior.

Key Concept

Properties and field deflection of natural radioactive emissions

Question 126Question →

A nuclear fusion reaction between a helium-3 nucleus (

{^{3}_{2}\text{He}}

) and a deuterium nucleus (

{^{2}_{1}\text{H}}

) produces a helium-4 nucleus (

{^{4}_{2}\text{He}}

) and a proton (

{^{1}_{1}\text{p}}

) according to the equation:

{^{3}_{2}\text{He}} + {^{2}_{1}\text{H}} \rightarrow {^{4}_{2}\text{He}} + {^{1}_{1}\text{p}}

Given the atomic masses:
- Mass of ${^{3}_{2}\text{He}} = 3.0160\text{ u}$
- Mass of ${^{2}_{1}\text{H}} = 2.0141\text{ u}$
- Mass of ${^{4}_{2}\text{He}} = 4.0026\text{ u}$
- Mass of ${^{1}_{1}\text{p}} = 1.0078\text{ u}$
- $1\text{ u} = 931\text{ MeV}$

What is the total energy released in this reaction?

18.34\text{ MeV}

0.0197\text{ MeV}

9.17\text{ MeV}

36.68\text{ MeV}

Show answer & explanation

Answer: $18.34\text{ MeV}$

Answer

The total energy released in the nuclear fusion reaction is

18.34\text{ MeV}

.

The total energy released in a nuclear fusion reaction is proportional to the mass defect between reactants and products. The sum of reactant masses is

5.0301\text{ u}

and product masses is

5.0104\text{ u}

, giving a mass defect

\Delta m = 0.0197\text{ u}

. Multiplying this mass defect by the conversion factor

931\text{ MeV/u}

yields

18.34\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants before fusion.

Total reactant mass =

3.0160\text{ u} + 2.0141\text{ u} = 5.0301\text{ u}

.

The initial mass is the sum of the helium-3 nucleus mass and deuterium nucleus mass.

2

Calculate the total mass of the reaction products.

Total product mass =

4.0026\text{ u} + 1.0078\text{ u} = 5.0104\text{ u}

.

The final mass is the sum of the helium-4 nucleus mass and proton mass.

3

Determine the mass defect (loss of mass during fusion).

\Delta m = 5.0301\text{ u} - 5.0104\text{ u} = 0.0197\text{ u}.

The mass defect represents the missing mass that is converted into energy.

4

Convert the mass defect into energy released using Einstein's energy equivalent constant.

Energy = 0.0197\text{ u} \times 931\text{ MeV/u} = 18.3407\text{ MeV} \approx 18.34\text{ MeV}.

Multiplying the mass defect in atomic mass units by

931\text{ MeV/u}

yields the energy released.

Key Concept

Mass defect and energy release in nuclear fusion

Question 127Question →

When a mixed beam containing $\alpha$ -particles, $\beta$ -particles, and $\gamma$ -rays passes through a uniform electric field, the $\beta$ -particles undergo a significantly greater lateral deflection than the $\alpha$ -particles. Which of the following best explains this observation?

The charge-to-mass ratio of a

\beta

-particle is significantly greater than that of an

\alpha

-particle.

The magnitude of the electric force acting on a

\beta

-particle is greater than that acting on an

\alpha

-particle.

\beta

-particles travel at a much lower velocity than

\alpha

-particles, allowing the electric field more time to act on them.

\alpha

-particles possess no net electric charge and are therefore unaffected by the electric field.

Show answer & explanation

Answer: The charge-to-mass ratio of a $\beta$ -particle is significantly greater than that of an $\alpha$ -particle.

Answer

The charge-to-mass ratio of a

\beta

-particle is significantly greater than that of an

\alpha

-particle.

In a uniform electric field

E

, a charged particle experiences an acceleration given by

a = \frac{qE}{m}

. The degree of deflection is directly proportional to this acceleration, and hence to the charge-to-mass ratio (

\frac{q}{m}

). Even though an

\alpha

-particle carries twice the magnitude of charge of a

\beta

-particle (

2e

compared to

e

), an

\alpha

-particle is about

7300

times more massive than a

\beta

-particle. Consequently, the charge-to-mass ratio of the

\beta

-particle is thousands of times larger, causing it to undergo a significantly greater deflection in the field.

Step-by-Step Solution

1

Identify the relationship between electric field deflection and particle properties.

Deflection in a transverse electric field is proportional to acceleration

a = \frac{qE}{m}

, meaning deflection depends on the specific charge (charge-to-mass ratio,

\frac{q}{m}

).

Newton's second law (

F = ma

) combined with Electrostatic force (

F = qE

) gives

a = \frac{qE}{m}

.

2

Compare the charges and masses of

\alpha

-particles and

\beta

-particles.

For an

\alpha

-particle (

^{4}_{2}\text{He}^{2+}

),

q_{\alpha} = +2e

and

m_{\alpha} \approx 7300 m_e

. For a

\beta

-particle (

^{0}_{-1}\text{e}

),

q_{\beta} = -e

and

m_{\beta} = m_e

.

An alpha particle consists of 2 protons and 2 neutrons, making it much more massive than a single electron (beta particle).

3

Calculate and compare the charge-to-mass ratios.

\left|\frac{q_{\beta}}{m_{\beta}}\right| = \frac{e}{m_e}

, while

\left|\frac{q_{\alpha}}{m_{\alpha}}\right| \approx \frac{2e}{7300 m_e} = \frac{e}{3650 m_e}

. Thus,

\left|\frac{q_{\beta}}{m_{\beta}}\right| \approx 3650 \times \left|\frac{q_{\alpha}}{m_{\alpha}}\right|

.

The vastly smaller mass of the beta particle dominates the ratio, resulting in a much larger charge-to-mass ratio and therefore a greater deflection.

Key Concept

Deflection of nuclear emissions in electric fields is governed by their charge-to-mass ratio (

\frac{q}{m}

).

Estimated Time:1m 15s

Question 128Question →

A medical sample contains a radioactive isotope with a half-life of $8\text{ days}$ . If the initial activity of the sample is $160\text{ kBq}$ , what is the activity corresponding to the portion of the sample that has decayed after $32\text{ days}$ ?

150\text{ kBq}

10\text{ kBq}

40\text{ kBq}

20\text{ kBq}

Show answer & explanation

Answer: $150\text{ kBq}$

Answer

150\text{ kBq}

In

32\text{ days}

, a sample with an

8\text{-day}

half-life undergoes

4

complete half-lives (

32 / 8 = 4

). The remaining activity is

160 \times (1/2)^4 = 10\text{ kBq}

. The activity that has decayed is the initial activity minus the remaining activity:

160\text{ kBq} - 10\text{ kBq} = 150\text{ kBq}

.

Step-by-Step Solution

1

Calculate the total number of half-lives (

n

) that have elapsed.

n = \frac{\text{Total time } (t)}{\text{Half-life } (T_{1/2})} = \frac{32\text{ days}}{8\text{ days}} = 4

Determining how many half-life periods fit into the total elapsed time.

2

Calculate the remaining activity (

A

) of the sample.

A = A_0 \left(\frac{1}{2}\right)^n = 160\text{ kBq} \times \left(\frac{1}{2}\right)^4 = 160 \times \frac{1}{16} = 10\text{ kBq}

Applying the radioactive decay law to find the undecayed fraction.

3

Calculate the activity corresponding to the decayed amount.

A_{\text{decayed}} = A_0 - A = 160\text{ kBq} - 10\text{ kBq} = 150\text{ kBq}

Subtracting the remaining activity from the initial activity gives the decayed activity.

Key Concept

Radioactive Decay Law and Half-life

Question 129Question →

A radioactive parent nucleus of Uranium-238 ( $^{238}_{92}\text{U}$ ) decays through a sequence of natural radioactive emissions to become a stable daughter nucleus of Lead-206 ( $^{206}_{82}\text{Pb}$ ). How many $\beta^-$ -particles are emitted in total during this decay process?

Show answer & explanation

Answer: 6

Answer

6

The decrease in mass number from

238

to

206

is

32

, which corresponds to

8

alpha emissions (

32 / 4 = 8

). Emitting

8

alpha particles reduces the atomic number from

92

to

76

(

92 - 16 = 76

). To reach the atomic number of Lead (

82

), the atomic number must be increased by

6

(

82 - 76 = 6

), which requires emitting

6

beta-minus particles.

Step-by-Step Solution

1

Calculate the total change in mass number and deduce the number of alpha particles emitted.

The total mass number decreases by

238 - 206 = 32

. Each alpha particle (

^{4}_{2}\text{He}

) decreases the mass number by

4

, while beta particles have a mass number of

0

. Therefore, the number of alpha particles is

\frac{32}{4} = 8

.

Beta particles carry negligible mass compared to nucleons, making alpha emission the sole mechanism for mass number reduction.

2

Calculate the intermediate atomic number after all alpha decays.

Each alpha particle removes

2

units of atomic number. The total reduction in atomic number due to

8

alpha particles is

8 \times 2 = 16

. The intermediate atomic number is

92 - 16 = 76

.

Alpha particles consist of 2 protons and 2 neutrons.

3

Determine the number of beta-minus particles required to reach the final atomic number.

The final atomic number of Lead-206 is

82

. The atomic number must increase by

82 - 76 = 6

. Since each

\beta^-

-particle increases the atomic number by

1

, exactly

6

beta-minus particles are emitted.

A beta-minus decay converts a neutron into a proton, increasing the atomic number by 1 without altering the mass number.

Key Concept

Balancing mass numbers and atomic numbers in radioactive decay series

Question 130Question →

A radioactive parent nucleus $^{A}_{Z}\text{X}$ emits one $\alpha$ -particle and two $\beta^-$ -particles during a natural radioactive decay process to form a daughter nucleus $\text{Y}$ . Which of the following correctly describes the relationship between the daughter nucleus $\text{Y}$ and the parent nucleus $\text{X}$ ?

It is an isotope of the parent element with a mass number that is 4 units lower than that of the parent nucleus.

It is an isobar of the parent element with an atomic number that is 2 units higher than that of the parent nucleus.

It is an isotope of the parent element with a mass number that is 2 units lower than that of the parent nucleus.

It is a different chemical element with an atomic number 2 units lower and a mass number 4 units lower.

Show answer & explanation

Answer: It is an isotope of the parent element with a mass number that is 4 units lower than that of the parent nucleus.

Answer

The daughter nucleus is an isotope of the parent element with a mass number that is 4 units lower than that of the parent nucleus.

Emitting one alpha particle reduces the nucleus mass number by 4 and atomic number by 2. The subsequent emission of two beta-minus particles increases the atomic number by 2 (1 for each beta particle) without affecting the mass number. Consequently, the net change returns the atomic number to its original value

Z

, while the mass number is reduced by 4. Nuclei with identical atomic numbers belong to the same element and are classified as isotopes.

Step-by-Step Solution

1

Analyze the change in mass number (

A

) and atomic number (

Z

) resulting from the emission of one

\alpha

-particle (

^{4}_{2}\text{He}

).

The mass number becomes

A - 4

and the atomic number becomes

Z - 2

.

An alpha particle consists of 2 protons and 2 neutrons, so emitting it removes 4 units of mass and 2 units of atomic charge.

2

Analyze the change in mass number (

A

) and atomic number (

Z

) resulting from the emission of two

\beta^-

-particles (

^{0}_{-1}\text{e}

).

The mass number remains

(A - 4)

and the atomic number becomes

(Z - 2) + 2 = Z

.

Each beta-minus particle is emitted when a neutron turns into a proton, increasing the atomic number by 1 while keeping the total mass number unchanged.

3

Compare the final nuclear numbers

(A - 4, Z)

of daughter nucleus

\text{Y}

with parent nucleus

(A, Z)

.

Daughter

\text{Y}

has the same atomic number

Z

as parent

\text{X}

, but its mass number is

A - 4

.

Nuclides with the same atomic number

Z

but different mass numbers

A

are isotopes of the same element.

Key Concept

Nuclear Decay Balancing and Isotopes

Estimated Time:1m 30s

Question 131Question →

During a natural radioactive decay series, a parent nucleus of Actinium-227 ( $^{227}_{89}\text{Ac}$ ) emits $5$ $\alpha$ -particles and $3$ $\beta^-$ -particles to reach a stable state. What is the atomic number ( $Z$ ) of the resulting stable daughter nucleus?

Show answer & explanation

Answer: 82

Answer

The atomic number of the resulting stable daughter nucleus is 82.

Each alpha particle emission decreases the nuclear charge (atomic number) by 2, so emitting 5 alpha particles reduces the atomic number by 10. Each beta-minus emission increases the nuclear charge by 1, so emitting 3 beta-minus particles increases the atomic number by 3. Starting with an initial atomic number of 89, the resulting atomic number is 89 - 10 + 3 = 82.

Step-by-Step Solution

1

Determine the atomic number reduction from alpha emissions.

5 alpha particles reduce the atomic number by 5 * 2 = 10.

An alpha particle (^4_2He) carries 2 protons, so each emission decreases Z by 2.

2

Determine the atomic number increase from beta-minus emissions.

3 beta-minus particles increase the atomic number by 3 * 1 = 3.

A beta-minus particle (^0_-1e) is emitted when a neutron converts to a proton, increasing Z by 1.

3

Apply the conservation of atomic number to find the final value.

Z_final = 89 - 10 + 3 = 82.

Subtracting the alpha contribution and adding the beta-minus contribution from the initial atomic number gives the daughter nucleus's atomic number.

Key Concept

Conservation of atomic number (charge) in natural radioactive decay chains

Question 132Question →

In a photoelectric effect experiment, monochromatic light of frequency $f$ (where $f > f_0$ ) illuminates a sodium metal surface, causing the emission of photoelectrons with a maximum kinetic energy $K_{\text{max}}$ and producing a saturation photoelectric current $I$ . If the intensity of the incident light is quadrupled while maintaining the frequency constant, what are the new values of the maximum kinetic energy of the photoelectrons and the saturation photoelectric current?

The maximum kinetic energy remains

K_{\text{max}}

, and the saturation current becomes

4I

.

The maximum kinetic energy becomes

4K_{\text{max}}

, and the saturation current remains

I

.

The maximum kinetic energy becomes

4K_{\text{max}}

, and the saturation current becomes

4I

.

The maximum kinetic energy remains

K_{\text{max}}

, and the saturation current remains

I

.

Show answer & explanation

Answer: The maximum kinetic energy remains $K_{\text{max}}$ , and the saturation current becomes $4I$ .

Answer

The maximum kinetic energy remains

K_{\text{max}}

, and the saturation current becomes

4I

.

According to Einstein's photoelectric theory, light intensity corresponds to the rate of photon arrival. Quadrupling the intensity quadruples the number of photons striking the surface per second, thereby quadrupling the rate of emitted photoelectrons and resulting in a saturation current of

4I

. However, individual photon energy depends strictly on frequency (

E = hf

). Because frequency remains constant, the maximum kinetic energy of the photoelectrons remains unchanged at

K_{\text{max}}

.

Step-by-Step Solution

1

Apply Einstein's photoelectric equation to analyze the maximum kinetic energy.

K_{\text{max}} = hf - W_0

. Since both frequency

f

and work function

W_0

remain constant,

K_{\text{max}}

remains unchanged.

Photon energy depends only on frequency (

E = hf

), so intensity changes do not affect individual photon energies or the kinetic energy of emitted photoelectrons.

2

Analyze the relationship between light intensity and photoelectric current.

Intensity

I_{\text{light}}

is proportional to the number of incident photons per second. Quadrupling intensity increases photon flux by a factor of 4, quadrupling photoelectron rate to

4I

.

Each photon causes the emission of one photoelectron (assuming 100% quantum efficiency), making photoelectric current directly proportional to light intensity.

Key Concept

Independence of photoelectron kinetic energy from light intensity and direct proportionality of photoelectric current to light intensity.

Question 133Question →

In a gas discharge tube operated at a very low pressure of approximately $0.01\text{ mmHg}$ , cathode rays are produced. Which of the following statements accurately describes a characteristic property of cathode rays?

They are electromagnetic waves of extremely high frequency.

They cast sharp shadows of opaque objects placed in their path.

They are deflected toward the negatively charged plate in an electric field.

They originate from the anode and travel toward the cathode.

Show answer & explanation

Answer: They cast sharp shadows of opaque objects placed in their path.

Answer

Cathode rays cast sharp shadows of opaque objects placed in their path because they travel in straight lines.

The correct option correctly states that cathode rays cast sharp shadows of opaque objects in their path. This phenomenon demonstrates that cathode rays travel in straight lines from the cathode.

Step-by-Step Solution

1

Identify the nature of cathode rays

Cathode rays are streams of electrons emitted from the cathode in a low-pressure discharge tube.

The high electric potential difference accelerates free electrons away from the cathode surface.

2

Analyze propagation behavior

Cathode rays travel normal to the cathode in straight lines.

Because they travel in straight lines, any opaque barrier (such as a metal cross) blocks the stream completely, casting a clear shadow on the fluorescent tube wall.

Key Concept

Properties of Cathode Rays

Question 134Question →

An X-ray tube operates at an accelerating potential difference of $25.0\text{ kV}$ . Calculate the maximum frequency of the emitted X-ray radiation in units of $10^{18}\text{ Hz}$ . (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ and elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ ).

Show answer & explanation

Answer: 6.03

Answer

The maximum frequency of the emitted X-rays is

6.03 \times 10^{18}\text{ Hz}

, which gives a value of

6.03

in units of

10^{18}\text{ Hz}

.

The maximum frequency of X-ray photons emitted occurs when an accelerating electron transfers all of its kinetic energy (

e V

) into a single photon (

h f_{\text{max}}

). Substituting

e = 1.60 \times 10^{-19}\text{ C}

,

V = 25,000\text{ V}

, and

h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}

gives

f_{\text{max}} = 6.03 \times 10^{18}\text{ Hz}

, which equals

6.03

in the requested unit of

10^{18}\text{ Hz}

.

Step-by-Step Solution

1

Convert potential difference from kilovolts to volts

V = 25.0\text{ kV} = 25,000\text{ V}

Standard SI units are required for calculations.

2

Determine maximum kinetic energy of the incident electrons

E_{\text{max}} = e V = 1.60 \times 10^{-19}\text{ C} \times 25,000\text{ V} = 4.00 \times 10^{-15}\text{ J}

The maximum photon energy produced equals the full kinetic energy acquired by an accelerated electron.

3

Calculate maximum frequency

f_{\text{max}}

using Duane-Hunt relation

f_{\text{max}} = \frac{E_{\text{max}}}{h} = \frac{4.00 \times 10^{-15}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.03 \times 10^{18}\text{ Hz}

According to the Duane-Hunt law,

e V = h f_{\text{max}}

.

Key Concept

Duane-Hunt Law and Maximum X-ray Frequency

Question 135Question →

An X-ray tube is operated at an accelerating potential difference of $15.0\text{ kV}$ . Assuming all the kinetic energy of an electron is transferred into a single photon upon collision with the target, what is the minimum cutoff wavelength ($\lambda_{\min}$) of the emitted X-rays?

(Take Planck's constant $h = 6.60 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , and elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ )

8.25 \times 10^{-11}\text{ m}

1.32 \times 10^{-14}\text{ m}

1.21 \times 10^{10}\text{ m}^{-1}

8.25 \times 10^{-8}\text{ m}

Show answer & explanation

Answer: $8.25 \times 10^{-11}\text{ m}$

Answer

The minimum cutoff wavelength of the emitted X-rays is

8.25 \times 10^{-11}\text{ m}

.

The correct answer is derived using Duane-Hunt's equation

\lambda_{\min} = \frac{hc}{eV}

. Substituting the known values (

h = 6.60 \times 10^{-34}\text{ J s}

,

c = 3.00 \times 10^8\text{ m s}^{-1}

,

e = 1.60 \times 10^{-19}\text{ C}

,

V = 15.0 \times 10^3\text{ V}

) gives

\lambda_{\min} = 8.25 \times 10^{-11}\text{ m}

.

Step-by-Step Solution

1

Identify given parameters and convert to SI units.

Accelerating potential

V = 15.0\text{ kV} = 15.0 \times 10^3\text{ V}

, Planck's constant

h = 6.60 \times 10^{-34}\text{ J s}

, speed of light

c = 3.00 \times 10^8\text{ m s}^{-1}

, charge of electron

e = 1.60 \times 10^{-19}\text{ C}

.

SI units are mandatory to ensure dimensional correctness during photon energy evaluation.

2

Apply the Duane-Hunt Law for the maximum photon energy and minimum wavelength.

E_{\max} = e V = \frac{h c}{\lambda_{\min}} \implies \lambda_{\min} = \frac{h c}{e V}

.

The shortest wavelength corresponds to maximum energy transfer when an electron yields all its kinetic energy to a single X-ray photon.

3

Substitute numerical values into the equation and calculate

\lambda_{\min}

.

$\lambda_{\min} = \frac{6.60 \times 10^{-34} \times 3.00 \times 10^8}{1.60 \times 10^{-19} \times 15.0 \times 10^3} = \frac{1.98 \times 10^{-25}}{2.40 \times 10^{-15}} = 8.25 \times 10^{-11}\text{ m}$.

Carrying out the arithmetic yields the exact Duane-Hunt minimum cutoff wavelength.

Key Concept

Duane-Hunt Law and Cutoff Wavelength in X-ray Production

Question 136Question →

The threshold wavelength for photoelectric emission from a metallic surface is $500\text{ nm}$ . What is the work function of the metal in electron-volts ( $\text{eV}$ )? [Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.0 \times 10^{8}\text{ m s}^{-1}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ ]

Show answer & explanation

Answer: 2.48

Answer

The work function of the metal is

2.48\text{ eV}

(or

2.475\text{ eV}

).

The work function

W_0

is calculated using

W_0 = \frac{hc}{\lambda_0}

. Substituting

h = 6.6 \times 10^{-34}\text{ J s}

,

c = 3.0 \times 10^8\text{ m s}^{-1}

, and

\lambda_0 = 500 \times 10^{-9}\text{ m}

gives

W_0 = 3.96 \times 10^{-19}\text{ J}

. Converting to electron-volts yields

\frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

, which rounds to

2.48\text{ eV}

.

Step-by-Step Solution

1

State the relationship between work function and threshold wavelength

W_0 = \frac{hc}{\lambda_0}

The work function is the minimum energy required to liberate an electron, which corresponds to the maximum wavelength (threshold wavelength

\lambda_0

) that can cause emission.

2

Substitute the physical constants and threshold wavelength to calculate

W_0

in joules

W_0 = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{500 \times 10^{-9}} = 3.96 \times 10^{-19}\text{ J}

Evaluating

hc / \lambda_0

yields the work function energy in standard SI units (Joules).

3

Convert the calculated work function into electron-volts

W_0 = \frac{3.96 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.475\text{ eV}

Dividing the energy in Joules by

1.6 \times 10^{-19}\text{ J/eV}

converts the value to electron-volts.

Key Concept

Work Function and Threshold Wavelength Relationship

Question 137Question →

Gases under atmospheric pressure are generally poor conductors of electricity. However, when the gas pressure inside a discharge tube is lowered to approximately $0.1\text{ mmHg}$ under a high potential difference, electrical conduction readily occurs. What is the primary physical reason for this increase in electrical conductivity at reduced pressures?

The mean free path of ions increases, enabling them to gain sufficient kinetic energy to produce collisional ionization.

The frequency of molecular collisions increases significantly, generating thermal ions spontaneously.

The gas density increases at low pressure, providing a greater number of charge carriers per unit volume.

Neutral gas molecules decompose spontaneously into electrons without requiring collision energy.

Show answer & explanation

Answer: The mean free path of ions increases, enabling them to gain sufficient kinetic energy to produce collisional ionization.

Answer

The increased conductivity at low pressures occurs because the mean free path of free ions increases, allowing them to accelerate under the high electric field and gain enough kinetic energy to cause ionization by collision with neutral gas atoms.

The correct option correctly states that reducing gas pressure increases the mean free path of ions. Under high applied potential differences, these ions travel longer distances uninterrupted, acquiring enough kinetic energy to ionize neutral gas molecules upon impact and causing electrical breakdown.

Step-by-Step Solution

1

Analyze gas density at reduced pressure

Lowering the gas pressure reduces the number density of molecules inside the discharge tube.

Gas pressure is directly proportional to molecule concentration.

2

Determine the impact on mean free path

The mean free path (average distance traveled between successive collisions) increases.

Fewer gas molecules per unit volume mean fewer obstructions per unit distance.

3

Relate mean free path to kinetic energy and ionization

Ions accelerate over longer distances, acquiring kinetic energy

K = qE \cdot d

exceeding the ionization potential of the gas, causing collisional ionization.

Sufficient kinetic energy allows colliding ions to knock electrons out of neutral atoms, generating avalanche conduction.

Key Concept

Collisional Ionization and Mean Free Path in Gas Discharge

Question 138Question →

In a standard Coolidge X-ray tube, more than 90% of the kinetic energy of the fast-moving electrons striking the target anode is converted directly into X-rays.

True

False

Show answer & explanation

Answer: False

Answer

False

The assertion is false because X-ray generation is a highly inefficient physical process. Less than 1% of the kinetic energy possessed by the high-speed electrons striking the target is converted into X-radiation. More than 99% of the kinetic energy is converted into heat, necessitating target materials with very high melting points (such as tungsten) embeddded in copper blocks for rapid heat conduction.

Step-by-Step Solution

1

Examine the energy transformation process during X-ray generation in a Coolidge tube.

Accelerated electrons collide with the heavy metal target anode, undergoing inelastic scattering and atomic interactions.

Electrons lose kinetic energy rapidly upon impact with the target material.

2

Evaluate the energy conversion ratio between radiation and heat.

Only approximately 0.2% to 1% of the total electron kinetic energy produces X-ray radiation (both Bremsstrahlung and characteristic emissions). The remaining 99%+ is converted into heat.

Most electron impacts only excite outer electron shells and cause lattice vibrations (thermal agitation) rather than inner-shell ionization or violent retardation.

Key Concept

Efficiency of X-ray production and heat generation at the target anode

Estimated Time:45s

Question 139Question →

A narrow beam of cathode rays is produced by accelerating electrons from rest through an unknown potential difference $V$ . The beam then enters a region containing mutually perpendicular uniform electric and magnetic fields of magnitudes $1.2 \times 10^4\text{ V/m}$ and $3.0 \times 10^{-3}\text{ T}$ , respectively. If the cathode rays pass through the crossed fields completely undeflected, what is the value of the accelerating potential difference $V$ ? [Take electron charge $e = 1.6 \times 10^{-19}\text{ C}$ and electron mass $m = 9.0 \times 10^{-31}\text{ kg}$ ]

45\text{ V}

90\text{ V}

180\text{ V}

7.2 \times 10^{-18}\text{ V}

Show answer & explanation

Answer: $45\text{ V}$

Answer

The accelerating potential difference

V

is

45\text{ V}

.

For cathode rays passing undeflected through crossed fields, the electric force equals the magnetic force (

eE = evB

), giving speed

v = E/B = 4.0 \times 10^6\text{ m/s}

. The work done by the electric field during acceleration from rest equals the kinetic energy:

eV = \frac{1}{2}mv^2

. Rearranging for

V

yields

V = \frac{m v^2}{2e} = \frac{(9.0 \times 10^{-31}) (4.0 \times 10^6)^2}{2 \times (1.6 \times 10^{-19})} = 45\text{ V}

. Therefore, the option stating

45\text{ V}

is correct.

Step-by-Step Solution

1

Calculate the velocity

v

of the cathode ray electrons using the undeflected condition in crossed fields

v = \frac{E}{B} = \frac{1.2 \times 10^4\text{ V/m}}{3.0 \times 10^{-3}\text{ T}} = 4.0 \times 10^6\text{ m/s}

When cathode rays pass undeflected through perpendicular electric and magnetic fields, the electric force

F_e = eE

balances the magnetic force

F_b = evB

.

2

Relate the kinetic energy gained by the electrons to the accelerating potential difference

V

e V = \frac{1}{2} m v^2 \implies V = \frac{m v^2}{2 e}

The work done by the accelerating potential difference equals the final kinetic energy of the electrons starting from rest.

3

Substitute the known physical quantities into the potential difference formula

V = \frac{(9.0 \times 10^{-31}\text{ kg}) \times (4.0 \times 10^6\text{ m/s})^2}{2 \times (1.6 \times 10^{-19}\text{ C})} = \frac{9.0 \times 10^{-31} \times 1.6 \times 10^{13}}{3.2 \times 10^{-19}} = 45\text{ V}

Evaluating the expression yields the required voltage

V = 45\text{ V}

.

Key Concept

Deflection of Cathode Rays in Crossed Fields and Energy Conversion in Cathode Ray Tubes

Question 140Question →

Match each discharge tube phenomenon or experimental setup in Column A with its correct physical description or property in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Faraday dark space matches the non-luminous region separating the negative glow from the positive column (

\approx 1\text{ mmHg}

); Crookes dark space matches the dark region around the cathode that expands at very low pressure (

\approx 0.01\text{ mmHg}

); Positive column matches the bright luminous stream extending toward the anode; Paddle wheel matches the demonstration of kinetic energy and mechanical momentum of cathode rays.

Each feature of discharge tubes corresponds to specific gas pressure stages and physical properties of cathode rays. Faraday dark space is the non-luminous region at moderate pressures; Crookes dark space expands near the cathode at very low pressure (

\approx 0.01\text{ mmHg}

); the positive column is the luminous glowing gas stream; and the paddle wheel experiment proves cathode rays possess mechanical momentum.

Step-by-Step Solution

1

Analyze discharge tube pressure stages.

At moderate pressures (

\approx 1\text{ mmHg}

), the Faraday dark space appears between the negative glow and the luminous positive column.

Ionization processes create distinct bright and dark regions based on electron acceleration distances between collisions.

2

Identify high-vacuum phenomena.

At very low pressures (

\approx 0.01\text{ mmHg}

), the Crookes dark space expands across almost the entire length of the discharge tube.

Fewer gas molecules allow electrons to travel longer distances without colliding, shifting ionization further down the tube.

3

Correlate mechanical experiments with cathode ray properties.

A paddle wheel mounted on glass rails rotates when hit by cathode rays, showing they possess momentum.

Electrons have mass

m

and velocity

v

, transferring kinetic energy and momentum (

p = mv

) upon impact.

Key Concept

Conduction of Electricity Through Gases and Cathode Rays

Atomic and Nuclear Physics

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