Atomic and Nuclear Physics Practice Questions — JAMB UTME

Question 1Question →

An electron in a hydrogen atom undergoes a transition from an energy state of $-1.51\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . What is 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}

.

The energy of an emitted photon during an atomic transition is given by

\Delta E = E_{\text{initial}} - E_{\text{final}}

. Substituting the given levels yields

\Delta 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.

E_i = -1.51\text{ eV}

and

E_f = -3.40\text{ eV}

The energy of the photon corresponds to the difference between these two levels.

2

Apply the energy conservation formula for atomic emission.

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

When an electron drops to a lower energy level, a photon carrying the lost energy is released.

3

Substitute the values and evaluate the difference.

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

Subtracting the negative lower energy value yields a positive photon energy.

Key Concept

Photon energy from atomic level transitions

Question 2Question →

In a Coolidge X-ray tube, increasing the current passing through the filament increases the intensity of the emitted X-rays without altering their penetrating power (hardness).

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. Filament current controls the rate of thermionic emission (X-ray intensity), whereas accelerating voltage controls the energy and penetrating power (X-ray hardness).

The statement is correct because filament current solely regulates the rate of thermionic emission, controlling the total number of electrons striking the target per second and thus the intensity of the X-ray beam. Penetrating power (hardness) depends on the accelerating potential difference, which dictates maximum electron kinetic energy and minimum X-ray wavelength.

Step-by-Step Solution

1

Analyze the effect of increasing filament current on electron emission in an X-ray tube.

Higher filament current causes greater heating, resulting in an increased rate of thermionic emission (more electrons emitted per second).

Thermionic emission depends directly on temperature, which increases with filament heating current.

2

Relate electron emission rate to X-ray intensity.

A higher number of incident electrons per second yields a proportional increase in the number of X-ray photons emitted per second (higher intensity).

Intensity measures the energy radiated per unit area per unit time, which depends on the total photon count.

3

Evaluate the factors determining X-ray penetrating power (hardness).

Penetrating power remains constant because the accelerating voltage

V

across the anode and cathode was not changed (

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

).

Photon energy and minimum cutoff wavelength are strictly determined by accelerating voltage, not by filament current.

Key Concept

Independence of X-ray Intensity and X-ray Hardness (Penetrating Power)

Question 3Question →

An electron in a hydrogen atom drops from an excited state with energy $-0.85\text{ eV}$ to a lower energy state of $-3.40\text{ eV}$ . What is the wavelength of the photon emitted during this transition? ( $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , $c = 3.0 \times 10^8\text{ m/s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

4.85 \times 10^{-7}\text{ m}

2.91 \times 10^{-7}\text{ m}

7.76 \times 10^{-26}\text{ m}

6.18 \times 10^{14}\text{ m}

Show answer & explanation

Answer: $4.85 \times 10^{-7}\text{ m}$

Answer

The wavelength of the emitted photon is

4.85 \times 10^{-7}\text{ m}

.

The energy of the emitted photon is given by

\Delta E = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

. Converting to Joules gives

2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

. Substituting into

\lambda = \frac{hc}{\Delta E}

yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{4.08 \times 10^{-19}} = 4.85 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the energy difference between the initial and final states

\Delta E = E_i - E_f = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

The energy of the emitted photon equals the energy difference between the two levels.

2

Convert the energy difference from electron-volts to Joules

\Delta E = 2.55 \times 1.6 \times 10^{-19}\text{ J} = 4.08 \times 10^{-19}\text{ J}

Standard SI units are required for calculations involving Planck's constant.

3

Calculate photon wavelength using

\lambda = \frac{hc}{\Delta E}

\lambda = \frac{(6.6 \times 10^{-34}\text{ J}\cdot\text{s})(3.0 \times 10^8\text{ m/s})}{4.08 \times 10^{-19}\text{ J}} = 4.85 \times 10^{-7}\text{ m}

Relating photon energy to wavelength using Planck's constant and speed of light.

Key Concept

Photon Emission and Atomic Transitions

Question 4Question →

In a cathode-ray experiment, a beam of electrons passes undeflected through mutually perpendicular electric and magnetic fields. If the electric field intensity 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 speed of the electrons in the beam?

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

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

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

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

Show answer & explanation

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

Answer

The speed of the electrons in the beam is

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

.

When a cathode ray beam passes undeflected through perpendicular electric (

E

) and magnetic (

B

) fields, the electric force (

eE

) and magnetic force (

evB

) are equal in magnitude and opposite in direction. Equating

eE = evB

simplifies 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}

yields an electron speed of

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

.

Step-by-Step Solution

1

Identify the condition for no deflection in crossed electric and magnetic fields.

The electrostatic force

F_e = eE

equals the magnetic force

F_b = evB

.

For zero net deflection, the opposing electric and magnetic forces must balance each other exactly.

2

Rearrange the force balance equation to solve for speed

v

.

v = \frac{E}{B}

Canceling the charge

e

on both sides of

eE = evB

isolates velocity

v

as the ratio of electric field to magnetic field.

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}

Evaluating

\frac{4.0}{2.0} \times 10^{4 - (-3)}

gives

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

.

Key Concept

Velocity selector condition for cathode rays in crossed electric and magnetic fields

Question 5Question →

A radioactive parent nucleus with an initial mass number of $226$ undergoes a natural decay sequence in which it emits $3$ $\alpha$ -particles and $2$ $\beta^-$ -particles. What is the mass number of the resulting daughter nucleus?

Show answer & explanation

Answer: 214

Answer

The mass number of the resulting daughter nucleus is 214.

Emitting an

\alpha

-particle reduces the nuclear mass number

A

by

4

, while emitting a

\beta^-

-particle leaves the mass number unchanged. Emitting

3

\alpha

-particles reduces the mass number by

3 \times 4 = 12

. Subtracting

12

from the original mass number of

226

yields a final mass number of

214

.

Step-by-Step Solution

1

Identify the mass number change caused by alpha and beta emissions.

Each alpha particle decreases the mass number by 4, while each beta particle causes no change in mass number.

An alpha particle consists of 2 protons and 2 neutrons (mass number 4), whereas a beta particle is an electron with negligible mass in nucleon terms (mass number 0).

2

Calculate the total change in mass number.

Total mass number reduction = 3 * 4 = 12.

Three alpha particles are emitted in the decay sequence.

3

Calculate the mass number of the daughter nucleus.

Daughter mass number = 226 - 12 = 214.

Subtracting the total change in mass number from the original parent mass number.

Key Concept

Conservation of mass number in radioactive decay series

Question 6Question →

In an X-ray tube, which of the following adjustments will increase the penetrating power (hardness) of the emitted X-rays?

Increasing the accelerating potential difference between the anode and cathode

Increasing the current passing through the filament heater

Decreasing the potential difference across the X-ray tube

Increasing the surface area of the target anode

Show answer & explanation

Answer: Increasing the accelerating potential difference between the anode and cathode

Answer

Increasing the accelerating potential difference between the anode and cathode increases the penetrating power of X-rays.

Increasing the accelerating potential difference between the anode and cathode increases the energy transferred to the colliding electrons. According to the Duane-Hunt law (

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

), higher potential difference leads to higher frequency X-ray photons, which increases their penetrating power (hardness).

Step-by-Step Solution

1

Identify the factor controlling electron kinetic energy

Maximum kinetic energy of electrons is

E_{\text{max}} = e V

, directly proportional to the tube voltage

V

.

Electrons gain energy from the electric field established by the potential difference.

2

Relate electron energy to X-ray photon energy and penetrating power

Higher kinetic energy produces X-ray photons of higher maximum frequency (

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

) and shorter minimum wavelength (hard X-rays).

Hard X-rays with higher frequencies have greater energy and higher penetrating power.

Key Concept

X-ray Hardness vs. Intensity Control

Estimated Time:45s

Question 7Question →

Monochromatic light of frequency $9.0 \times 10^{14}\text{ Hz}$ is incident on a clean potassium emitter surface having a threshold frequency of $5.0 \times 10^{14}\text{ Hz}$ . Given that Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ , what is the stopping potential in volts required to completely arrest the photoelectric current?

Show answer & explanation

Answer: 1.65

Answer

The stopping potential required to arrest the emitted photoelectrons is

1.65\text{ V}

.

Applying Einstein's photoelectric equation

K_{\max} = h f - h f_0 = h(f - f_0)

gives a maximum kinetic energy of

2.64 \times 10^{-19}\text{ J}

. Dividing this by the electronic charge

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

yields the stopping potential

V_s = 1.65\text{ V}

.

Step-by-Step Solution

1

Determine the net energy available for photoelectron kinetic energy using Einstein's photoelectric equation

K_{\max} = h(f - f_0) = 6.6 \times 10^{-34}\text{ J}\cdot\text{s} \times (9.0 \times 10^{14} - 5.0 \times 10^{14})\text{ Hz} = 2.64 \times 10^{-19}\text{ J}

Photoelectron emission occurs only when photon energy

hf

exceeds the work function

W_0 = h f_0

, with the excess energy appearing as maximum kinetic energy.

2

Express stopping potential in terms of electron charge and maximum kinetic energy

V_s = \frac{K_{\max}}{e} = \frac{2.64 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ C}} = 1.65\text{ V}

The stopping potential

V_s

does work

e V_s

equal to the maximum kinetic energy

K_{\max}

of the fastest photoelectrons to bring them to rest.

Key Concept

Einstein's Photoelectric Equation and Stopping Potential Relationship

Question 8Question →

Radioactive nuclei emit different types of emissions ( $\alpha$ , $\beta^{-}$ , $\beta^{+}$ , and $\gamma$ ) characterized by distinct deflection behaviors, energy spectra, and physical properties when passing through electric fields. Match each type of radiation emission listed on the left with its corresponding physical properties and field response on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Alpha particle emission matches weak deflection towards the negative electrode with discrete energy levels; Beta-minus particle emission matches strong deflection towards the positive electrode with a continuous spectrum; Beta-plus particle emission matches strong deflection towards the negative electrode accompanied by a neutrino; Gamma ray emission matches zero deflection in electric fields and high speed.

Matching each emission type correctly relies on evaluating electric field deflection (determined by charge sign and

q/m

ratio) and spectral characteristics. Alpha particles are heavy positive ions showing slight deflection toward the negative electrode with discrete energy states. Beta-minus emissions are light negative particles deflecting strongly toward the positive electrode in a continuous spectrum. Beta-plus emissions are light positive particles deflecting strongly toward the negative electrode alongside a neutrino. Gamma radiation is uncharged photon energy showing zero deflection.

Step-by-Step Solution

1

Analyze charge and mass ratios of radioactive emissions to determine magnetic and electric field deflection direction and magnitude.

Alpha particles (

+2e

, mass

4\text{ u}

) deflect slightly toward negative plate; Beta-minus (

-e

, negligible mass) deflects strongly toward positive plate; Beta-plus (

+e

, negligible mass) deflects strongly toward negative plate; Gamma rays (

0

charge,

0

mass) do not deflect.

Deflection angle in an electric field depends directly on the charge-to-mass ratio (

q/m

) and the sign of the charge.

2

Evaluate energy spectra characteristics and secondary particle emissions for decay modes.

Alpha decay produces discrete kinetic energy peaks. Beta decay produces a continuous spectrum due to three-body decay sharing kinetic energy with a neutrino or antineutrino. Gamma photons carry discrete transition energy.

Conservation of momentum and energy in three-body beta decay requires kinetic energy sharing with the (anti)neutrino.

3

Pair each radiation type with its full physical description.

Alpha matches weak deflection to negative plate with discrete energy; Beta-minus matches strong deflection to positive plate with continuous spectrum; Beta-plus matches strong deflection to negative plate with neutrino co-emission; Gamma matches no deflection and lowest specific ionization.

Combines field deflection, charge-to-mass ratio, and spectral traits into unique matching pairings.

Key Concept

Deflection characteristics, charge-to-mass ratios, and energy spectrum nature of alpha, beta, and gamma radiation emissions

Question 9Question →

An X-ray tube operates at an accelerating potential of $50.0\text{ kV}$ . Given Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ , the speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , and the elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ , what is the minimum wavelength $\lambda_{\min}$ of the emitted continuous X-rays, and how does this cutoff wavelength respond to an increase in the anode potential?

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

, and

\lambda_{\min}

decreases when the anode potential is increased.

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

, and

\lambda_{\min}

decreases when the anode potential is increased.

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

, and

\lambda_{\min}

remains unchanged because

\lambda_{\min}

depends only on the filament current.

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

, and

\lambda_{\min}

increases when the anode potential is increased.

Show answer & explanation

Answer: $2.49 \times 10^{-11}\text{ m}$ , and $\lambda_{\min}$ decreases when the anode potential is increased.

Answer

The minimum wavelength is

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

, and

\lambda_{\min}

decreases when the anode potential is increased.

By the Duane-Hunt law, the maximum photon energy produced by electron impact equals the kinetic energy of the incident electron:

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

. Substituting

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

,

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

,

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

, and

V = 5.00 \times 10^4\text{ V}

yields

\lambda_{\min} = 2.49 \times 10^{-11}\text{ m}

. Because

\lambda_{\min}

is inversely proportional to

V

, increasing the anode potential decreases the minimum wavelength, making the beam more penetrating (harder).

Step-by-Step Solution

1

Convert the operating voltage to standard SI units (volts).

V = 50.0\text{ kV} = 50.0 \times 10^3\text{ V} = 5.00 \times 10^4\text{ V}

.

Equations involving fundamental constants require input quantities in SI base units.

2

Apply the Duane-Hunt law for maximum photon energy / minimum wavelength in continuous X-ray production.

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

.

The maximum kinetic energy of an accelerating electron is completely converted into a single photon of minimum wavelength.

3

Substitute the physical values into the minimum wavelength formula.

$\lambda_{\min} = \frac{(6.63 \times 10^{-34}\text{ J s})(3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C})(5.00 \times 10^4\text{ V})} = \frac{1.989 \times 10^{-25}}{8.00 \times 10^{-15}} = 2.48625 \times 10^{-11}\text{ m} \approx 2.49 \times 10^{-11}\text{ m}$.

Direct numerical evaluation yields the minimum cutoff wavelength.

4

Analyze the functional relationship between anode potential

V

and cutoff wavelength

\lambda_{\min}

.

Since

\lambda_{\min} \propto \frac{1}{V}

, an increase in

V

results in a decrease in

\lambda_{\min}

.

Higher potential imparts greater kinetic energy to striking electrons, enabling emission of higher-frequency (shorter-wavelength) photons.

Key Concept

Duane-Hunt Law and Control Parameters of X-ray Production

Estimated Time:2m 0s

Question 10Question →

An atom in a gas discharge tube has a ground state energy level of $-10.4\text{ eV}$ and an excited energy level of $-3.8\text{ eV}$ . An electron absorbs a single photon to undergo a transition directly from the ground state to this excited level. If the frequency of the absorbed photon is expressed as $x \times 10^{15}\text{ Hz}$ , what is the numerical value of $x$ ? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 1.6

Answer

The numerical value of

x

is 1.6.

The energy of the absorbed photon is equal to the difference between the excited state and ground state energies:

\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}

. Converting this to Joules yields

6.6 \times 1.6 \times 10^{-19}\text{ J} = 1.056 \times 10^{-18}\text{ J}

. Using Einstein's photon relation

E = hf

, the frequency is

f = \frac{1.056 \times 10^{-18}}{6.6 \times 10^{-34}} = 1.6 \times 10^{15}\text{ Hz}

, which gives

x = 1.6

.

Step-by-Step Solution

1

Determine the energy absorbed during transition

\Delta E = 6.6\text{ eV}

The energy of the absorbed photon equals the energy difference between the initial and final states:

\Delta E = -3.8\text{ eV} - (-10.4\text{ eV}) = 6.6\text{ eV}

.

2

Convert energy from eV to Joules

\Delta E = 1.056 \times 10^{-18}\text{ J}

Multiply by

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

to convert energy to standard SI units.

3

Calculate photon frequency

f = 1.6 \times 10^{15}\text{ Hz}

Use Planck's energy equation

f = \frac{\Delta E}{h} = \frac{1.056 \times 10^{-18}\text{ J}}{6.6 \times 10^{-34}\text{ J}\cdot\text{s}} = 1.6 \times 10^{15}\text{ Hz}

.

Key Concept

Photon energy and atomic transition frequency

Estimated Time:2m 0s

Question 11Question →

Monochromatic light with a photon energy of $4.5\text{ eV}$ strikes the surface of a sodium plate inside a vacuum tube. If the work function of sodium is $2.1\text{ eV}$ , what is the maximum kinetic energy of the emitted photoelectrons in electron-volts ( $\text{eV}$ )?

Show answer & explanation

Answer: 2.4

Answer

The maximum kinetic energy of the emitted photoelectrons is 2.4 eV.

According to Einstein's photoelectric theory, energy is conserved such that the energy of an incident photon (

E

) is partly used to liberate an electron from the metal surface (work function

W_0

) and the remaining energy appears as the maximum kinetic energy (

K_{\text{max}}

) of the photoelectron. Subtracting

2.1\text{ eV}

from

4.5\text{ eV}

gives

2.4\text{ eV}

.

Step-by-Step Solution

1

Identify the given values from the problem statement

Photon energy

E = 4.5\text{ eV}

, Work function

W_0 = 2.1\text{ eV}

Establishing known variables helps determine the correct photoelectric relation to apply.

2

Apply Einstein's photoelectric equation

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

The maximum kinetic energy of photoelectrons equals the excess energy of incident photons after overcoming the work function.

3

Calculate the value

K_{\text{max}} = 4.5\text{ eV} - 2.1\text{ eV} = 2.4\text{ eV}

Subtracting the work function from the photon energy yields the final numerical answer.

Key Concept

Einstein's Photoelectric Equation and Energy Conservation

Question 12Question →

In a Coolidge X-ray tube, electrons are accelerated from rest through an unknown potential difference $V$ towards a target metal. If the shortest cutoff wavelength of the resulting continuous X-ray spectrum is recorded as $0.0414\text{ nm}$ , what is the operating potential difference $V$ of the tube?

(Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J s}$ , speed of light $c = 3.00 \times 10^8\text{ m s}^{-1}$ , elementary charge $e = 1.60 \times 10^{-19}\text{ C}$ )

30.0\text{ kV}

15.0\text{ kV}

300\text{ V}

60.0\text{ kV}

Show answer & explanation

Answer: $30.0\text{ kV}$

Answer

The operating potential difference

V

of the tube is

30.0\text{ kV}

.

According to the Duane-Hunt law for X-ray emission, the shortest wavelength

\lambda_{\min}

corresponds to the maximum kinetic energy acquired by an electron accelerated through a voltage

V

. The mathematical relationship is

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

. Substituting

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

,

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

,

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

, and

\lambda_{\min} = 4.14 \times 10^{-11}\text{ m}

gives

V = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30.0\text{ kV}

.

Step-by-Step Solution

1

Convert the given minimum wavelength from nanometers to meters

\lambda_{\min} = 0.0414\text{ nm} = 0.0414 \times 10^{-9}\text{ m} = 4.14 \times 10^{-11}\text{ m}

SI unit consistency requires length to be in meters.

2

Apply the Duane-Hunt law for continuous X-ray production

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

The maximum photon energy equals the kinetic energy of the incident electron.

3

Rearrange the equation to solve for the accelerating voltage

V

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

Isolating the operating potential difference

V

on one side of the equation.

4

Substitute the physical constants and calculate

V

V = \frac{(6.63 \times 10^{-34}\text{ J s}) (3.00 \times 10^8\text{ m s}^{-1})}{(1.60 \times 10^{-19}\text{ C}) (4.14 \times 10^{-11}\text{ m})} = \frac{1.989 \times 10^{-25}}{6.624 \times 10^{-30}} = 30,012\text{ V} \approx 30.0\text{ kV}

Performing numeric evaluation gives the operating potential difference in kilovolts.

Key Concept

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

Estimated Time:2m 0s

Question 13Question →

Which of the following emissions resulting from natural radioactivity passes through a strong electric field without undergoing any deflection?

Gamma rays (

\gamma

-rays)

Alpha particles (

\alpha

-particles)

Beta-minus particles (

\beta^-

-particles)

Positrons (

\beta^+

-particles)

Show answer & explanation

Answer: Gamma rays ( $\gamma$ -rays)

Answer

Gamma rays (

\gamma

-rays) carry no electrical charge, so they pass through an electric field without any deflection.

Gamma rays consist of neutral electromagnetic radiation (photons). Because their charge is zero, they experience no electrostatic force in an electric field and continue in a straight, undeflected path.

Step-by-Step Solution

1

Identify the nature and electric charge of each type of radioactive emission.

Alpha particles have a charge of

+2e

, beta particles have a charge of

-e

or

+e

, and gamma rays are high-energy photons with zero net electric charge.

An electric field exerts an electrostatic force

F = qE

only on particles that carry a non-zero electric charge

q

.

2

Determine which emission experiences zero electrostatic force.

Since

q = 0

for gamma rays,

F = 0

, meaning gamma rays experience no deflection.

Uncharged radiation travels in a straight line unaffected by electrostatic or magnetic fields.

Key Concept

Electric Field Deflection of Radioactive Emissions

Estimated Time:45s

Question 14Question →

A radioactive isotope has a half-life of $4\text{ hours}$ . If a sample initially contains $80\text{ g}$ of the isotope, what mass of the isotope, in grams, will remain undecayed after $12\text{ hours}$ ?

Show answer & explanation

Answer: 10

Answer

The mass of the radioactive isotope remaining undecayed after

12\text{ hours}

is

10\text{ g}

.

After

3

half-lives (

12\text{ hours}

total elapsed time with a half-life of

4\text{ hours}

), the fraction of the initial sample remaining is

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

. Multiplying this fraction by the initial mass of

80\text{ g}

gives

10\text{ g}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives (

n

)

n = \frac{12\text{ hours}}{4\text{ hours}} = 3

half-lives

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

2

Calculate the mass remaining after

3

half-lives

N = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ g}

The remaining mass halves during each half-life interval according to the exponential decay rule

N = N_0 (1/2)^n

.

Key Concept

Radioactive Decay Law and Half-life

Question 15Question →

A heavy radioactive parent nucleus undergoes a decay series emitting five $\alpha$ -particles and four $\beta^{-}$ -particles to form a stable daughter nucleus. What is the net change in the mass number ( $A$ ) and atomic number ( $Z$ ) of the nucleus?

Mass number decreases by

20

, and atomic number decreases by

6

Mass number decreases by

20

, and atomic number decreases by

14

Mass number decreases by

9

, and atomic number decreases by

6

Mass number decreases by

20

, and atomic number decreases by

10

Show answer & explanation

Answer: Mass number decreases by $20$ , and atomic number decreases by $6$

Answer

The mass number decreases by

20

, and the atomic number decreases by

6

.

An alpha particle emission removes 2 protons and 2 neutrons (

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

), reducing the atomic mass number by 4 and the atomic number by 2. Five alpha emissions reduce

A

by

20

and

Z

by

10

. A beta-minus emission (

^{0}_{-1}e

) converts a neutron into a proton, leaving

A

unchanged while increasing

Z

by 1. Four beta-minus emissions increase

Z

by 4. Combining these gives a net decrease in mass number of 20 and a net decrease in atomic number of 6.

Step-by-Step Solution

1

Determine the change in mass number (

A

) and atomic number (

Z

) due to

\alpha

-emissions.

Each

\alpha

-particle (

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

) decreases

A

by

4

and

Z

by

2

. For

5

\alpha

-particles:

\Delta A_{\alpha} = 5 \times (-4) = -20

,

\Delta Z_{\alpha} = 5 \times (-2) = -10

.

Alpha particles consist of two protons and two neutrons.

2

Determine the change in mass number (

A

) and atomic number (

Z

) due to

\beta^{-}

-emissions.

Each

\beta^{-}

-particle (

^{0}_{-1}e

) leaves

A

unchanged and increases

Z

by

1

. For

4

\beta^{-}

-particles:

\Delta A_{\beta} = 0

,

\Delta Z_{\beta} = 4 \times (+1) = +4

.

Beta-minus decay involves a neutron converting into a proton, emitting an electron.

3

Sum the total changes for

A

and

Z

.

Net

\Delta A = -20 + 0 = -20

(decrease by

20

). Net

\Delta Z = -10 + 4 = -6

(decrease by

6

).

Combining independent particle emission effects yields the total net change.

Key Concept

Natural radioactive radiation emissions change nuclear composition: alpha emission (

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

) reduces

A

by

4

and

Z

by

2

, whereas beta-minus emission (

^{0}_{-1}e

) keeps

A

constant and increases

Z

by

1

.

Question 16Question →

Match each atomic model with its key defining feature or experimental foundation.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Thomson Model matches with electrons embedded in a positive sphere; Rutherford Model matches with the dense positive nucleus discovered via alpha particle scattering; Bohr Model matches with quantized non-radiating energy levels; Quantum Wave Model matches with standing matter waves in electron orbits.

Each atomic model corresponds directly to its historic milestone: Thomson proposed electrons scattered inside a uniform sphere of positive charge; Rutherford discovered the compact positive nucleus through alpha scattering experiments; Bohr introduced quantized stationary states to account for emission spectral lines; and the Quantum Wave Model integrated de Broglie matter wave harmonics into orbital mechanics.

Step-by-Step Solution

1

Identify Thomson's Plum Pudding Model features

Describes electrons distributed inside a broad, uniform cloud/sphere of positive charge.

Formulated after J.J. Thomson discovered the electron before the nucleus was known.

2

Identify Rutherford's Nuclear Model features

Demonstrated that positive charge is concentrated in a tiny nucleus because alpha particles bounced back at large angles.

Geiger-Marsden alpha scattering experiment disproved the uniform charge distribution model.

3

Identify Bohr's Model features

Postulated quantized non-radiating stationary orbits where angular momentum is restricted to integral multiples of

h / 2\pi

.

Successfully explained the discrete wavelengths of the hydrogen emission spectrum.

4

Identify the Quantum Wave Model features

Explains quantization by treating orbiting electrons as de Broglie standing waves around the nucleus.

Orbits are stable only when an integral number of electron wavelengths fit along the orbital circumference.

Key Concept

Characteristics and experimental foundations of atomic models

Question 17Question →

An electron in an atom transitions from an excited state with energy $-1.7 \times 10^{-19}\text{ J}$ to a lower state with energy $-5.0 \times 10^{-19}\text{ J}$ . What is the wavelength of the emitted photon in nanometers ( $\text{nm}$ )? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ and the speed of light $c = 3.0 \times 10^8\text{ m/s}$ )

Show answer & explanation

Answer: 600

Answer

The wavelength of the emitted photon is 600 nm.

The energy released during the atomic transition is

\Delta E = (-1.7 \times 10^{-19}\text{ J}) - (-5.0 \times 10^{-19}\text{ J}) = 3.3 \times 10^{-19}\text{ J}

. Substituting this into

\lambda = \frac{hc}{\Delta E}

yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{3.3 \times 10^{-19}} = 6.0 \times 10^{-7}\text{ m}

. Converting to nanometers (

1\text{ m} = 10^9\text{ nm}

) gives

600\text{ nm}

.

Step-by-Step Solution

1

Calculate energy of the emitted photon

\Delta E = 3.3 \times 10^{-19}\text{ J}

The energy of the photon equals the difference between the initial higher energy level and the final lower energy level.

2

Calculate wavelength in meters using Planck's relation

\lambda = 6.0 \times 10^{-7}\text{ m}

Rearranging

\Delta E = \frac{hc}{\lambda}

gives

\lambda = \frac{hc}{\Delta E}

.

3

Convert wavelength to nanometers

600\text{ nm}

Multiply by

10^9\text{ nm/m}

to obtain the final value in nanometers.

Key Concept

Energy level transitions and photon emission wavelength

Question 18Question →

In a photoelectric effect experiment, monochromatic light with a frequency greater than the threshold frequency of a cesium metal emitter is directed at the surface. If the intensity of the incident light is doubled while keeping its frequency constant, which of the following statements is correct?

The rate of emitted photoelectrons doubles, but their maximum kinetic energy remains unchanged.

Both the maximum kinetic energy of the photoelectrons and the rate of emission double.

The maximum kinetic energy doubles, while the rate of photoelectron emission remains unchanged.

The maximum kinetic energy increases, but the rate of photoelectron emission decreases.

Show answer & explanation

Answer: The rate of emitted photoelectrons doubles, but their maximum kinetic energy remains unchanged.

Answer

The rate of emitted photoelectrons doubles, but their maximum kinetic energy remains unchanged.

According to Einstein's photoelectric equation,

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

, the maximum kinetic energy of emitted photoelectrons depends strictly on the frequency of incident light and the work function of the metal emitter. Doubling light intensity at constant frequency increases the number of incident photons per second, which proportionally doubles the number of photoelectrons ejected per second (photoelectric current) without affecting their kinetic energy.

Step-by-Step Solution

1

Relate light intensity to the rate of incident photons.

Light intensity

I

is directly proportional to the photon flux (number of photons per unit area per unit time).

Intensity is defined by total photon energy per unit area per second (

I = N h f / (A t)

).

2

Determine the impact of intensity on the rate of photoelectron emission.

Doubling intensity doubles the number of photons hitting the surface per second, thereby doubling the photoelectron emission rate.

Photoelectric emission follows a 1-to-1 photon-to-electron collision process.

3

Apply Einstein's photoelectric equation to analyze kinetic energy.

The maximum kinetic energy

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

remains constant.

Since frequency

f

and work function

W_0

are kept constant, the energy per photon

hf

and maximum kinetic energy

K_{\text{max}}

are unaffected by light intensity.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 19Question →

A particle of mass $6.63 \times 10^{-27}\text{ kg}$ has a de Broglie wavelength of $2.0 \times 10^{-13}\text{ m}$ . What is the speed of the particle in $\text{m/s}$ ? (Take Planck's constant $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ )

Show answer & explanation

Answer: 500000

Answer

The speed of the particle is

500,000\text{ m/s}

(or

5.0 \times 10^5\text{ m/s}

).

According to de Broglie's hypothesis, the matter wavelength

\lambda

of a particle is related to its momentum

p = m v

by

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

. Rearranging for speed yields

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

. Substituting

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

,

m = 6.63 \times 10^{-27}\text{ kg}

, and

\lambda = 2.0 \times 10^{-13}\text{ m}

gives

v = \frac{6.63 \times 10^{-34}}{6.63 \times 10^{-27} \times 2.0 \times 10^{-13}} = 500,000\text{ m/s}

.

Step-by-Step Solution

1

Identify the given physical quantities and formula

m = 6.63 \times 10^{-27}\text{ kg}

,

\lambda = 2.0 \times 10^{-13}\text{ m}

,

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

. The de Broglie equation is

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

.

The de Broglie wavelength formula relates matter wave properties to particle momentum.

2

Isolate the target variable (speed

v

)

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

Algebraic manipulation is needed to solve directly for speed.

3

Substitute parameters and compute the result

v = \frac{6.63 \times 10^{-34}}{(6.63 \times 10^{-27})(2.0 \times 10^{-13})} = \frac{6.63 \times 10^{-34}}{1.326 \times 10^{-39}} = 500,000\text{ m/s}

Performing scientific notation division yields the numerical particle speed.

Key Concept

de Broglie Wavelength and Particle Speed

Question 20Question →

Monochromatic light of frequency $8.0 \times 10^{14}\text{ Hz}$ illuminates a metal surface with a work function of $2.2\text{ eV}$ . If the intensity of the incident light is doubled while keeping the frequency constant, what is the maximum kinetic energy of the emitted photoelectrons? ( $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

1.1\text{ eV}

2.2\text{ eV}

3.3\text{ eV}

4.4\text{ eV}

Show answer & explanation

Answer: $1.1\text{ eV}$

Answer

The maximum kinetic energy of the emitted photoelectrons remains

1.1\text{ eV}

.

According to Einstein's photoelectric equation,

K_{\max} = hf - W_0

. For incident light of frequency

8.0 \times 10^{14}\text{ Hz}

, the photon energy is

3.3\text{ eV}

. Subtracting the work function of

2.2\text{ eV}

yields

K_{\max} = 1.1\text{ eV}

. Because maximum kinetic energy depends solely on photon frequency and the metal's work function, changing the light intensity has no impact on

K_{\max}

.

Step-by-Step Solution

1

Calculate the energy of the incident photon in Joules and convert to electron-volts (eV).

E = hf = (6.6 \times 10^{-34}\text{ J}\cdot\text{s}) \times (8.0 \times 10^{14}\text{ Hz}) = 5.28 \times 10^{-19}\text{ J} = \frac{5.28 \times 10^{-19}}{1.6 \times 10^{-19}}\text{ eV} = 3.3\text{ eV}

.

Einstein's photoelectric equation requires knowing the energy of each incident photon.

2

Apply Einstein's photoelectric equation to calculate the maximum kinetic energy (

K_{\max}

).

K_{\max} = E - W_0 = 3.3\text{ eV} - 2.2\text{ eV} = 1.1\text{ eV}

.

The work function (

W_0

) is the minimum energy needed to liberate an electron from the metal surface.

3

Analyze the effect of doubling light intensity at constant frequency.

K_{\max}

remains

1.1\text{ eV}

.

Light intensity is proportional to the number of photons striking the surface per second, affecting the rate of electron emission (photocurrent), but not the energy of individual photons or the maximum kinetic energy of the emitted electrons.

Key Concept

Independence of photoelectron maximum kinetic energy from light intensity

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