Atomic and Nuclear Physics Practice Questions — JAMB UTME — Page 2

Question 21Question →

A narrow beam containing $\alpha$ -particles, $\beta^-$ -particles, and $\gamma$ -rays enters a region of uniform magnetic field acting perpendicularly into the plane of the paper. If all three emissions enter with the same speed, which of the following observations correctly describes their paths in the magnetic field?

The

\beta^-

-particles curve more sharply than the

\alpha

-particles and in the opposite direction, while

\gamma

-rays continue straight without deflection.

The

\alpha

-particles curve more sharply than the

\beta^-

-particles in the opposite direction, while

\gamma

-rays are deflected parallel to the magnetic field.

Both

\alpha

-particles and

\beta^-

-particles curve in the same direction, but the

\alpha

-particles undergo a larger deflection due to their

+2e

charge.

The

\gamma

-rays curve towards the negative region while

\alpha

-particles and

\beta^-

-particles travel straight through without deflection.

Show answer & explanation

Answer: The $\beta^-$ -particles curve more sharply than the $\alpha$ -particles and in the opposite direction, while $\gamma$ -rays continue straight without deflection.

Answer

The beta-minus particles curve more sharply than the alpha particles and in the opposite direction, while gamma rays continue straight without deflection.

The magnetic force on a moving charged particle supplies the necessary centripetal force (

qvB = \frac{mv^2}{r}

), yielding a radius of curvature of

r = \frac{mv}{qB}

. Because a beta-minus particle has an extremely small mass compared to an alpha particle, its mass-to-charge ratio is much lower, resulting in a much tighter (sharper) circular arc. Because alpha particles are positively charged and beta-minus particles are negatively charged, they experience forces in opposite directions. Gamma rays have no charge and are therefore unaffected by the magnetic field.

Step-by-Step Solution

1

Identify the charge and mass characteristics of each emission type

Alpha particles (

\alpha

) have charge

+2e

and large mass (

\approx 4\text{ u}

). Beta-minus particles (

\beta^-

) have charge

-1e

and extremely small mass (

\approx 0.00055\text{ u}

). Gamma rays (

\gamma

) carry zero charge and zero rest mass.

Magnetic forces depend on moving charge (

F = qvB

), while inertia against trajectory changes depends on mass.

2

Determine deflection directions for each emission

Alpha particles (positive) and beta-minus particles (negative) deflect in opposite directions. Gamma rays (neutral) experience zero magnetic force and travel in a straight line.

Particles with opposite sign charges experience magnetic Lorentz forces in opposite directions.

3

Compare curvature extent using the magnetic radius formula

r = \frac{mv}{qB}

The mass-to-charge ratio

\frac{m}{q}

for

\beta^-

is far smaller than that for

\alpha

. Consequently,

r_{\beta} \ll r_{\alpha}

, meaning the

\beta^-

trajectory forms a tighter, sharper curve.

A smaller radius of curvature corresponds to a sharper trajectory deflection in the magnetic field.

Key Concept

Deflection and Curvature of Radioactive Radiation in Magnetic Fields

Question 22Question →

A sample of a radioactive element has a half-life of $3\text{ hours}$ . If the initial mass of the sample is $48\text{ g}$ , what mass of the element has decayed after a period of $9\text{ hours}$ ?

42\text{ g}

6\text{ g}

12\text{ g}

24\text{ g}

Show answer & explanation

Answer: $42\text{ g}$

Answer

The mass of the element that has decayed after 9 hours is

42\text{ g}

.

In a period of 9 hours, exactly 3 half-lives pass (

9/3 = 3

). The fraction of the radioactive substance remaining is

(1/2)^3 = 1/8

, which corresponds to

48\text{ g} \times (1/8) = 6\text{ g}

. Therefore, the mass that has decayed is

48\text{ g} - 6\text{ g} = 42\text{ g}

.

Step-by-Step Solution

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{9\text{ hours}}{3\text{ hours}} = 3

Dividing the total elapsed time by the half-life period gives the number of decay cycles.

2

Calculate the mass of the sample remaining (

N

) after 3 half-lives.

N = N_0 \left(\frac{1}{2}\right)^n = 48\text{ g} \times \left(\frac{1}{2}\right)^3 = 48\text{ g} \times \frac{1}{8} = 6\text{ g}

The remaining quantity decreases by a factor of 2 for each half-life.

3

Calculate the mass of the sample that has decayed.

\text{Mass decayed} = N_0 - N = 48\text{ g} - 6\text{ g} = 42\text{ g}

The amount decayed is equal to the initial mass minus the remaining mass.

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:1m 30s

Question 23Question →

A radioactive detector records an initial disintegration rate of $6400\text{ counts per minute}$ from a freshly prepared isotope. If the half-life of the isotope is $5\text{ hours}$ , determine the total time, in hours, required for the count rate to decrease to $400\text{ counts per minute}$ .

Show answer & explanation

Answer: 20

Answer

The total time required for the disintegration rate to decrease to

400\text{ counts per minute}

is

20\text{ hours}

.

The fraction of activity remaining is

\frac{400}{6400} = \frac{1}{16}

. Expressing this as a power of one-half,

\left(\frac{1}{2}\right)^4 = \frac{1}{16}

, shows that

4

half-lives have elapsed. Multiplying

4

half-lives by

5\text{ hours}

per half-life yields a total duration of

20\text{ hours}

.

Step-by-Step Solution

1

Calculate the ratio of remaining activity to initial activity

\frac{N}{N_0} = \frac{400}{6400} = \frac{1}{16}

To find the fraction of the original radioactive substance that remains undecayed.

2

Determine the number of elapsed half-lives

n = 4

Since \left(\frac{1}{2}\right)^n = \frac{1}{16} = \left(\frac{1}{2}\right)^4, four complete half-lives have passed.

3

Compute the total elapsed time

t = 4 \times 5 = 20\text{ hours}

Total time equals the number of half-lives multiplied by the duration of one half-life.

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:1m 30s

Question 24Question →

The mass of a helium nucleus $^{4}_{2}\text{He}$ is $4.0015\text{ u}$ . If the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the binding energy per nucleon of the helium nucleus? (Take $1\text{ u} = 931\text{ MeV}$ )

7.10\text{ MeV}

28.40\text{ MeV}

14.20\text{ MeV}

0.0305\text{ MeV}

Show answer & explanation

Answer: $7.10\text{ MeV}$

Answer

The binding energy per nucleon of the helium nucleus is

7.10\text{ MeV}

.

The correct response accurately calculates the mass defect of the helium-4 nucleus (

0.0305\text{ u}

), converts it to total binding energy (

28.40\text{ MeV}

), and divides by the mass number

A = 4

to obtain

7.10\text{ MeV}

per nucleon.

Step-by-Step Solution

1

Calculate the total mass of the individual constituent nucleons (2 protons and 2 neutrons).

Total constituent mass =

2(1.0073\text{ u}) + 2(1.0087\text{ u}) = 2.0146\text{ u} + 2.0174\text{ u} = 4.0320\text{ u}

.

Helium-4 consists of

Z = 2

protons and

N = A - Z = 4 - 2 = 2

neutrons.

2

Calculate the mass defect (

\Delta m

).

\Delta m = 4.0320\text{ u} - 4.0015\text{ u} = 0.0305\text{ u}

.

Mass defect is the difference between the total mass of individual nucleons and the actual nuclear mass.

3

Convert mass defect into total binding energy (

E_b

).

E_b = 0.0305\text{ u} \times 931\text{ MeV/u} = 28.3955\text{ MeV}

.

Applying mass-energy equivalence using the factor

1\text{ u} = 931\text{ MeV}

.

4

Divide total binding energy by the total number of nucleons (

A = 4

).

\text{Binding Energy per nucleon} = \frac{28.3955\text{ MeV}}{4} = 7.098875\text{ MeV} \approx 7.10\text{ MeV}.

Binding energy per nucleon measures average stability per nuclear particle.

Key Concept

Binding Energy per Nucleon and Mass Defect

Estimated Time:1m 30s

Question 25Question →

A radioactive isotope has a half-life of $5\text{ days}$ . If the initial activity of a sample of this isotope is $400\text{ Bq}$ , what is the activity of the portion of the sample that has decayed after $15\text{ days}$ ?

350\text{ Bq}

50\text{ Bq}

100\text{ Bq}

300\text{ Bq}

Show answer & explanation

Answer: $350\text{ Bq}$

Answer

350\text{ Bq}

After 3 half-lives (15 days), one-eighth of the original activity remains, which equals 50 Bq. Therefore, seven-eighths of the original activity has decayed, giving 400 Bq - 50 Bq = 350 Bq.

Step-by-Step Solution

1

Calculate the number of half-lives that elapse in 15 days

Number of half-lives,

n = \frac{15\text{ days}}{5\text{ days}} = 3

Dividing the total time elapsed by the half-life period yields the number of decay half-lives.

2

Calculate the remaining activity of the sample

Remaining activity

A = A_0 \left(\frac{1}{2}\right)^n = 400 \times \left(\frac{1}{2}\right)^3 = 400 \times \frac{1}{8} = 50\text{ Bq}

The remaining fraction after

n

half-lives is

(1/2)^n

of the initial activity.

3

Subtract the remaining activity from the initial activity to find the decayed activity

Decayed activity

A_{\text{decayed}} = A_0 - A = 400\text{ Bq} - 50\text{ Bq} = 350\text{ Bq}

The portion that has decayed is equal to the total initial activity minus the activity that remains.

Key Concept

Distinction between remaining activity and decayed activity in radioactive decay calculations

Question 26Question →

A radioactive sample has an initial activity of $80\text{ Bq}$ and a half-life of $4\text{ days}$ . What is the activity of the sample that has decayed after $12\text{ days}$ ?

70\text{ Bq}

10\text{ Bq}

40\text{ Bq}

20\text{ Bq}

Show answer & explanation

Answer: $70\text{ Bq}$

Answer

The activity that has decayed after 12 days is

70\text{ Bq}

.

After 12 days, which equals 3 half-lives (

12 / 4 = 3

), the remaining activity of the sample is

80 \times (1/2)^3 = 10\text{ Bq}

. Consequently, the activity that has decayed is the initial activity minus the remaining activity:

80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives

n = \frac{t}{T_{1/2}} = \frac{12\text{ days}}{4\text{ days}} = 3\text{ half-lives}

Dividing the total time elapsed by the half-life period gives the number of decay cycles.

2

Calculate the remaining activity

A_{\text{remaining}} = A_0 \left(\frac{1}{2}\right)^n = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ Bq}

The remaining quantity decreases by half for each half-life cycle.

3

Calculate the decayed activity

A_{\text{decayed}} = A_0 - A_{\text{remaining}} = 80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}

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

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:45s

Question 27Question →

The atomic mass of a lithium nucleus $^{7}_{3}\text{Li}$ is $7.0160\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the lithium nucleus? ( $1\text{ u} = 931\text{ MeV}$ )

37.89\text{ MeV}

5.41\text{ MeV}

36.59\text{ MeV}

0.041\text{ MeV}

Show answer & explanation

Answer: $37.89\text{ MeV}$

Answer

The total binding energy of the lithium nucleus is

37.89\text{ MeV}

.

The correct answer is derived by finding the mass defect

\Delta m = [3(1.0073) + 4(1.0087)] - 7.0160 = 0.0407\text{ u}

and multiplying by

931\text{ MeV/u}

to obtain

37.89\text{ MeV}

.

Step-by-Step Solution

1

Determine the number of protons and neutrons in

^{7}_{3}\text{Li}

Z = 3

protons,

N = A - Z = 7 - 3 = 4

neutrons

The mass number

A=7

and atomic number

Z=3

define the nuclear composition.

2

Calculate total mass of individual nucleons

Mass of nucleons

= 3(1.0073\text{ u}) + 4(1.0087\text{ u}) = 3.0219\text{ u} + 4.0348\text{ u} = 7.0567\text{ u}

Summing the masses of constituent protons and neutrons.

3

Calculate mass defect

(\Delta m)

\Delta m = 7.0567\text{ u} - 7.0160\text{ u} = 0.0407\text{ u}

Mass defect is the difference between total constituent mass and nuclear mass.

4

Convert mass defect to energy

E_b = 0.0407\text{ u} \times 931\text{ MeV/u} = 37.8917\text{ MeV} \approx 37.89\text{ MeV}

Using the equivalence

1\text{ u} = 931\text{ MeV}

.

Key Concept

Mass Defect and Nuclear Binding Energy

Question 28Question →

In a nuclear fusion reaction, two light nuclei fuse together to form a heavier nucleus. The total mass of the reactants before fusion is $4.028\text{ u}$ , and the total mass of the products after fusion is $4.003\text{ u}$ . Given that $1\text{ u} = 931.5\text{ MeV}$ , what is the total energy released in this reaction in $\text{MeV}$ ?

Show answer & explanation

Answer: 23.2875

Answer

The total energy released in the reaction is 23.2875 MeV.

The energy released in nuclear fusion is calculated using the mass defect $\Delta m = m_{\text{reactants}} - m_{\text{products}}$. Subtracting

4.003\text{ u}

from

4.028\text{ u}

yields a mass defect of

0.025\text{ u}

. Multiplying this mass defect by

931.5\text{ MeV/u}

gives the total energy released as

23.2875\text{ MeV}

.

Step-by-Step Solution

1

Calculate the mass defect (\Delta m)

\Delta m = 4.028\text{ u} - 4.003\text{ u} = 0.025\text{ u}

Mass defect is the loss of mass during nuclear fusion that gets converted into energy.

2

Convert mass defect into energy released

E = 0.025\text{ u} \times 931.5\text{ MeV/u} = 23.2875\text{ MeV}

According to mass-energy equivalence, each atomic mass unit (u) of missing mass yields 931.5 MeV of energy.

Key Concept

Mass Defect and Energy Release in Nuclear Reactions

Question 29Question →

A radioactive sample has a half-life of $12\text{ minutes}$ . If its initial activity is $96\text{ Bq}$ , what is the remaining activity of the sample, in $\text{Bq}$ , after an elapsed time of $36\text{ minutes}$ ?

Show answer & explanation

Answer: 12

Answer

The remaining activity of the radioactive sample after

36\text{ minutes}

is

12\text{ Bq}

.

In

36\text{ minutes}

, exactly

3

half-lives elapse (

36 / 12 = 3

). The remaining activity reduces to

(1/2)^3 = 1/8

of the initial value, giving

96\text{ Bq} / 8 = 12\text{ Bq}

.

Step-by-Step Solution

1

Calculate the number of half-lives that have elapsed

n = 3

Divide the total elapsed time (

36\text{ minutes}

) by the half-life duration (

12\text{ minutes}

).

2

Calculate the remaining activity of the isotope

A = 12\text{ Bq}

After

n = 3

half-lives, the remaining activity fraction is

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

. Multiplying initial activity

96\text{ Bq}

by

\frac{1}{8}

yields

12\text{ Bq}

.

Key Concept

Radioactive Decay Law and Half-life

Question 30Question →

Match each type of natural radioactive emission with its corresponding characteristic physical property or behavior in electric and magnetic fields.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alpha emission matches strong ionizing power and low penetration (stopped by paper); Beta-minus emission matches deflection towards the positive electric plate; Gamma emission matches zero deflection and requirement of dense shielding like thick lead.

Alpha emissions possess strong ionizing capability and minimal penetration. Beta-minus emissions are negatively charged, causing them to deflect toward the positive plate in an electric field. Gamma emissions carry no charge and have the greatest penetrating power.

Step-by-Step Solution

1

Determine the physical nature and ionization capacity of Alpha particles.

Alpha particles (

^4_2\text{He}

) carry a charge of

+2e

and possess high mass. This yields intense ionization along their paths and low penetrating ability.

Massive double-charged particles undergo frequent ionizing collisions with gas molecules.

2

Analyze the electrostatic deflection behavior of Beta-minus particles.

Beta particles (

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

) carry a negative charge, so Coulomb forces pull them toward the positive potential plate.

Like charges repel and opposite electrical charges attract.

3

Analyze the electrostatic field interaction and penetrating ability of Gamma rays.

Gamma emissions are high-frequency photons with zero net charge and zero rest mass, passing straight without deviation and requiring heavy shielding such as lead.

Neutral photons do not experience electrostatic forces.

Key Concept

Distinct physical properties, ionizing powers, penetrating abilities, and field deflections of natural radioactive emissions.

Question 31Question →

An X-ray tube used in medical diagnostic imaging produces continuous X-rays with a minimum cutoff wavelength of $0.04125\text{ nm}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , the speed of light $c = 3.0 \times 10^{8}\text{ m/s}$ , and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the accelerating potential difference across the tube in kilovolts (kV)?

Show answer & explanation

Answer: 30

Answer

The accelerating potential difference across the X-ray tube is 30 kV.

By Duane-Hunt's law, the maximum energy of an emitted X-ray photon corresponds to the complete conversion of an electron's kinetic energy acquired across potential difference

V

:

e V = \frac{h c}{\lambda_{\text{min}}}

. Substituting

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

,

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

,

e = 1.6 \times 10^{-19}\text{ C}

, and

\lambda_{\text{min}} = 4.125 \times 10^{-11}\text{ m}

yields

V = 30,000\text{ V} = 30\text{ kV}

.

Step-by-Step Solution

1

Convert the given wavelength into SI units of meters

\lambda_{\text{min}} = 4.125 \times 10^{-11}\text{ m}

Standard SI units are required for calculations involving physical constants.

2

State the Duane-Hunt relation for maximum photon energy

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

The maximum photon energy occurs when all kinetic energy of an accelerating electron is converted into a single X-ray photon.

3

Rearrange the equation and substitute the numerical values to solve for

V

V = \frac{(6.6 \times 10^{-34})(3.0 \times 10^{8})}{(1.6 \times 10^{-19})(4.125 \times 10^{-11})} = 30,000\text{ V}

Evaluates the required accelerating potential difference in volts.

4

Express the final voltage in kilovolts (kV)

V = 30\text{ kV}

The question asks for the answer specifically in kilovolts.

Key Concept

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

Question 32Question →

In a cathode-ray experiment, a beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field intensity between the deflection plates is $4.0 \times 10^{4}\text{ V m}^{-1}$ and the magnetic flux density is $2.0 \times 10^{-3}\text{ T}$ , what is the velocity of the electrons in the cathode-ray beam?

2.0 \times 10^{7}\text{ m s}^{-1}

5.0 \times 10^{-8}\text{ m s}^{-1}

8.0 \times 10^{1}\text{ m s}^{-1}

6.0 \times 10^{7}\text{ m s}^{-1}

Show answer & explanation

Answer: $2.0 \times 10^{7}\text{ m s}^{-1}$

Answer

The velocity of the electrons in the cathode-ray beam is

2.0 \times 10^{7}\text{ m s}^{-1}

.

For cathode rays passing undeflected through perpendicular electric (

E

) and magnetic (

B

) fields, the electric force

eE

pulling electrons toward the positive plate is exactly balanced by the magnetic force

evB

. Equating

eE = evB

leads directly to

v = \frac{E}{B}

. Substituting

E = 4.0 \times 10^{4}\text{ V m}^{-1}

and

B = 2.0 \times 10^{-3}\text{ T}

gives

v = \frac{4.0 \times 10^{4}}{2.0 \times 10^{-3}} = 2.0 \times 10^{7}\text{ m s}^{-1}

.

Step-by-Step Solution

1

Identify the equilibrium condition for an undeflected electron beam in crossed fields.

The electric force

F_e = eE

balances the magnetic Lorentz force

F_b = evB

, so

eE = evB

.

When cathode rays pass undeflected, the net transverse force acting on each electron is zero.

2

Rearrange the equilibrium equation to solve for velocity

v

.

v = \frac{E}{B}

Canceling the elementary charge

e

from both sides isolates electron speed as a function of field strengths.

3

Substitute the given values into the velocity equation.

v = \frac{4.0 \times 10^{4}\text{ V m}^{-1}}{2.0 \times 10^{-3}\text{ T}} = 2.0 \times 10^{7}\text{ m s}^{-1}

Executing the division yields the velocity of cathode rays.

Key Concept

Velocity selector principle in crossed electric and magnetic fields (J.J. Thomson cathode ray velocity determination)

Question 33Question →

A metal surface inside a vacuum cell is illuminated by incident photons each having an energy of $6.2\text{ eV}$ . If the maximum kinetic energy of the emitted photoelectrons is $2.4\text{ eV}$ , what is the work function of the metal in electron-volts ( $\text{eV}$ )?

Show answer & explanation

Answer: 3.8

Answer

The work function of the metal is

3.8\text{ eV}

.

According to Einstein's photoelectric equation, the total energy of an incident photon (

E = 6.2\text{ eV}

) equals the work function of the metal (

W_0

) plus the maximum kinetic energy of the ejected photoelectrons (

K_{\max} = 2.4\text{ eV}

). Subtracting the kinetic energy from the photon energy gives

W_0 = 6.2\text{ eV} - 2.4\text{ eV} = 3.8\text{ eV}

.

Step-by-Step Solution

1

State Einstein's photoelectric equation

E = W_0 + K_{\max}

Relates the incident photon energy to the metal work function and photoelectron kinetic energy.

2

Rearrange the equation to solve for the work function

W_0 = E - K_{\max}

Isolates the work function

W_0

on one side of the equation.

3

Substitute the given numerical values and compute

W_0 = 6.2\text{ eV} - 2.4\text{ eV} = 3.8\text{ eV}

Evaluates the difference to obtain the minimum energy needed to remove an electron.

Key Concept

Einstein's Photoelectric Equation and Work Function

Question 34Question →

Match each experimental observation of cathode rays in a discharge tube with the corresponding physical property or characteristic it demonstrates.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1. Formation of a sharp shadow of a Maltese cross matches with Cathode rays travel in straight lines. 2. Rotation of a paddle wheel matches with Cathode rays possess particle mass and mechanical momentum. 3. Deflection toward a positively charged plate matches with Cathode rays carry a negative electrical charge. 4. Deflection perpendicular to a magnetic field matches with Cathode rays act as a current of moving charged particles obeying magnetic force laws.

Each experimental setup provides specific proof of a cathode ray property: sharp shadow formation proves straight-line motion; turning a paddle wheel demonstrates particle mass and momentum; attraction to a positive plate confirms negative charge; and deflection in a magnetic field confirms that the beam acts as moving electrical charges.

Step-by-Step Solution

1

Analyze the Maltese cross shadow experiment

Sharp shadows indicate straight-line propagation of rays from the cathode surface.

Light and particle beams traveling in straight lines produce sharp geometrical shadows of opaque obstructions.

2

Analyze the paddle wheel experiment

The paddle wheel rotates when struck by cathode rays, demonstrating kinetic energy and momentum transfer.

Mechanical rotation requires a force resulting from the momentum transfer of moving material particles.

3

Analyze electric field deflection

The ray path bends toward the positive anode plate.

Electrostatic attraction pulls negatively charged entities toward positive electric potentials.

4

Analyze magnetic field deflection

The beam bends laterally perpendicular to both the trajectory and the magnetic field vector.

Moving electrical charges experience a magnetic Lorentz force given by

F = qvB\sin\theta

.

Key Concept

Experimental evidence establishing the properties of cathode rays

Estimated Time:1m 30s

Question 35Question →

Increasing the intensity of incident monochromatic light of a fixed frequency above the threshold frequency increases the maximum kinetic energy of the photoelectrons emitted from a metal surface.

True

False

Show answer & explanation

Answer: False

Answer

The statement is False. Increasing light intensity at constant frequency increases the number of emitted photoelectrons per second (photoelectric current), but leaves their maximum kinetic energy unchanged.

The statement is false. In quantum physics, increasing the intensity of monochromatic light increases the photon flux, which elevates the rate of photoelectron emission (photoelectric current). However, the maximum kinetic energy of each photoelectron depends solely on the energy of an individual photon (

E = hf

) minus the metal's work function (

W_0

), both of which remain unchanged when intensity is increased at fixed frequency.

Step-by-Step Solution

1

Identify the factors determining maximum kinetic energy in the photoelectric effect.

Maximum kinetic energy

K_{\text{max}}

is governed by Einstein's photoelectric equation

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

.

Energy transfer occurs on a one-photon-to-one-electron basis.

2

Determine the physical quantity affected by changing light intensity.

Intensity dictates the number of photons striking the metal surface per unit time.

Higher intensity means a greater photon flux, which increases the photoelectron emission rate.

3

Evaluate the statement.

The statement falsely attributes an increase in kinetic energy to an increase in intensity.

Since photon frequency

f

and work function

W_0

remain constant,

K_{\text{max}}

does not change.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 36Question →

Match each feature observed during electrical discharge through a gas with its corresponding pressure stage or physical characteristic.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Cathode Glow matches 'Luminous glow appearing right next to the cathode at around 10 mmHg pressure', Striations match 'Alternating bright and dark bands in the positive column at around 1 mmHg pressure', and Crookes Dark Space matches 'Dark region extending to fill most of the tube at very low pressure around 0.01 mmHg'.

Each feature of gas discharge corresponds to a specific pressure regime inside the discharge tube: Cathode Glow occurs at ~10 mmHg, Striations form at ~1 mmHg in the positive column, and Crookes Dark Space expands to cover most of the tube at ~0.01 mmHg where cathode rays are freely emitted.

Step-by-Step Solution

1

Identify the discharge stage for Cathode Glow.

Cathode glow occurs at moderate pressure (~10 mmHg) directly adjacent to the negative electrode.

Initial gas ionization near the cathode causes luminescence at this pressure stage.

2

Identify the discharge stage for Striations.

Striations represent light and dark divisions of the positive column at ~1 mmHg.

Periodic ionization and recombination along the tube produce alternating luminous discs.

3

Identify the discharge stage for Crookes Dark Space.

Crookes dark space expands as pressure drops to ~0.01 mmHg.

At very low pressures, electrons travel longer distances without colliding, causing the dark space to fill most of the discharge tube.

Key Concept

Stages of electric conduction through gases at reduced pressure.

Question 37Question →

In a hydrogen atom modeled according to Bohr's postulates, an electron undergoes a transition from an initial excited state $n$ to the ground state $n=1$ , emitting a photon whose wavelength is $\lambda = \frac{16}{15R}$ , where $R$ is the Rydberg constant. What is the orbital angular momentum of the electron in its initial excited state prior to the transition?

\frac{h}{2\pi}

\frac{h}{\pi}

\frac{2h}{\pi}

\frac{4h}{\pi}

Show answer & explanation

Answer: $\frac{2h}{\pi}$

Answer

\frac{2h}{\pi}

Using the Rydberg formula

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right)

with

\lambda = \frac{16}{15R}

, we find

\frac{15}{16} = 1 - \frac{1}{n^2}

, which gives

n = 4

. By Bohr's angular momentum quantization condition,

L = \frac{nh}{2\pi} = \frac{4h}{2\pi} = \frac{2h}{\pi}

.

Step-by-Step Solution

1

Apply the Rydberg formula for hydrogen to determine the principal quantum number

n

of the initial state.

\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{n^2} \right) \implies \frac{15R}{16} = R \left( 1 - \frac{1}{n^2} \right) \implies 1 - \frac{1}{n^2} = \frac{15}{16} \implies \frac{1}{n^2} = \frac{1}{16} \implies n = 4

.

The Rydberg formula links the photon wavelength emitted during a transition between energy levels to the principal quantum numbers.

2

Calculate the orbital angular momentum

L

for

n=4

using Bohr's quantization condition.

L = \frac{n h}{2\pi} = \frac{4 h}{2\pi} = \frac{2 h}{\pi}

.

Bohr's quantization postulate dictates that orbital angular momentum is an integral multiple of

\frac{h}{2\pi}

.

Key Concept

Bohr's Atomic Model: Spectral transitions and angular momentum quantization

Estimated Time:2m 0s

Question 38Question →

During Rutherford's alpha particle scattering experiment, the vast majority of alpha particles passed straight through the gold foil with little to no deflection. What conclusion about atomic structure was drawn directly from this specific observation?

The atom consists mostly of empty space

The positive charge is uniformly distributed throughout the atom like a sphere

Electrons orbit the nucleus in fixed, quantized energy levels

Alpha particles are absorbed by electrons, causing them to jump to higher energy states

Show answer & explanation

Answer: The atom consists mostly of empty space

Answer

The atom consists mostly of empty space

In Rutherford's alpha particle scattering experiment, the fact that over 99% of alpha particles passed straight through the foil undeflected indicates that they encountered no massive positive obstacle. This directly proved that the volume of an atom consists predominantly of empty space, with positive charge concentrated in a tiny central region called the nucleus.

Step-by-Step Solution

1

Analyze the experimental observation from Rutherford's experiment

The vast majority of positively charged alpha particles suffered no deflection when passing through gold foil.

Deflection occurs when alpha particles pass close to a dense positive charge.

2

Deduce the physical implication for the structure of the atom

Since almost no deflection occurred for most particles, they encountered no heavy concentration of positive charge or mass in their path.

This proves that the dense positively charged nucleus occupies an extremely small fraction of the total atomic volume, leaving the rest as empty space.

Key Concept

Rutherford's Alpha Scattering Experiment and Nuclear Atom Model

Question 39Question →

Match each historical atomic model on the left with its defining postulate, experimental outcome, or theoretical limitation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thomson's model matches the diffuse positive sphere disproved by alpha particle backscattering; Rutherford's model matches the dense nucleus with classical radiation collapse limitations; Bohr's model matches quantized angular momentum in non-radiating orbits; Sommerfeld's extension matches elliptical sub-shells and relativistic adjustments for fine-structure splitting.

Each model directly maps to its defining theoretical contribution or failure mechanism: Thomson's diffuse charge sphere failed under

\alpha

-particle scattering; Rutherford's nuclear atom suffered from classical radiation instability; Bohr's model introduced angular momentum quantization

L = n\hbar

; and Sommerfeld's model extended orbits to ellipses with relativistic velocity corrections to account for fine structure.

Step-by-Step Solution

1

Analyze Thomson's Plum Pudding Model

Thomson proposed electrons embedded in a sea of positive charge. This continuous distribution could not account for

\alpha

-particles rebounding at angles greater than

90^\circ

.

Identify the historical assumption and experimental contradiction for Thomson's model.

2

Analyze Rutherford's Planetary Model

Rutherford deduced a concentrated positive core (nucleus). However, according to Maxwellian electrodynamics, orbiting electrons accelerate continuously, radiating energy until collapsing into the nucleus.

Identify the primary theoretical failure of classical planetary electron orbits.

3

Analyze Bohr's Quantized Model

Bohr introduced the non-classical postulate that electrons exist in stable stationary states with angular momentum

L = \frac{nh}{2\pi}

, accurately yielding the Rydberg formula for hydrogen.

Recognize the quantum postulate resolving Rutherford's radiation collapse.

4

Analyze Sommerfeld's Extension

To explain fine spectral line splitting not accounted for by circular Bohr orbits, Sommerfeld introduced elliptical orbits with azimuthal quantum numbers and relativistic mass variation at high electron velocities.

Connect fine-structure spectral features to relativistic elliptical orbital modifications.

Key Concept

Development and Limitations of Historical Atomic Models

Question 40Question →

In a hydrogen atom modeled according to Bohr's theory, an electron undergoes a transition from an excited state with an energy of $-1.51\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . Calculate the energy of the emitted photon in electron-volts ( $\text{eV}$ ).

Show answer & explanation

Answer: 1.89

Answer

The energy of the emitted photon is

1.89\text{ eV}

.

According to Bohr's atomic model, when an electron drops from an initial higher energy level

E_i

to a final lower energy level

E_f

, a photon is emitted carrying energy

E = E_i - E_f

. Substituting the given values gives

E = -1.51\text{ eV} - (-3.40\text{ eV}) = 1.89\text{ eV}

.

Step-by-Step Solution

1

Identify the initial and final energy states of the electron.

E_i = -1.51\text{ eV}

and

E_f = -3.40\text{ eV}

.

The electron moves from a higher (less negative) energy state to a lower (more negative) energy state.

2

Apply Bohr's energy quantization formula for photon emission

E_{\text{photon}} = E_i - E_f

.

E_{\text{photon}} = -1.51 - (-3.40) = 1.89\text{ eV}

.

By energy conservation, the energy lost by the transitioning electron equals the energy of the emitted photon.

Key Concept

Bohr's Energy Transition Postulate

Estimated Time:1m 0s

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