Atomic and Nuclear Physics Practice Questions — JAMB UTME — Page 4

Question 61Question →

An electron beam traveling horizontally enters a region containing mutually perpendicular uniform electric and magnetic fields. The electric field intensity is $4.4 \times 10^3\text{ V m}^{-1}$ and the magnetic flux density is $5.0 \times 10^{-4}\text{ T}$ . If the beam passes through undeflected and subsequently enters a region containing only the magnetic field $B$ , what is the radius of the circular path executed by the electrons? (Take the specific charge of an electron $\frac{e}{m} = 1.76 \times 10^{11}\text{ C kg}^{-1}$ )

0.10\text{ m}

0.88\text{ m}

10.0\text{ m}

0.20\text{ m}

Show answer & explanation

Answer: $0.10\text{ m}$

Answer

The radius of the circular path executed by the electrons in the magnetic field is

0.10\text{ m}

.

Under velocity selection conditions, equal electric and magnetic forces (

eE = evB

) establish electron speed

v = \frac{E}{B} = 8.8 \times 10^6\text{ m s}^{-1}

. When moving solely through magnetic field

B

, magnetic force provides centripetal force (

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

), yielding orbital radius

r = \frac{v}{(e/m)B} = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = 0.10\text{ m}

.

Step-by-Step Solution

1

Calculate the electron velocity using the undeflected crossed fields condition (velocity selector)

v = \frac{E}{B} = \frac{4.4 \times 10^3\text{ V m}^{-1}}{5.0 \times 10^{-4}\text{ T}} = 8.8 \times 10^6\text{ m s}^{-1}

When electric and magnetic forces are equal and opposite (

eE = evB

), the electrons travel in a straight line undeflected.

2

Equate magnetic force to centripetal force in the region with magnetic field only

evB = \frac{mv^2}{r} \implies r = \frac{mv}{eB} = \frac{v}{\left(\frac{e}{m}\right)B}

The magnetic Lorentz force provides the necessary centripetal force for circular motion.

3

Substitute numerical values to find the radius

r

r = \frac{8.8 \times 10^6}{1.76 \times 10^{11} \times 5.0 \times 10^{-4}} = \frac{8.8 \times 10^6}{8.8 \times 10^7} = 0.10\text{ m}

Direct algebraic simplification yields the orbital radius in meters.

Key Concept

Deflection of cathode rays (electrons) in crossed electric and magnetic fields (velocity selector) and circular orbital dynamics in uniform magnetic fields.

Estimated Time:2m 0s

Question 62Question →

In Rutherford's $\alpha$ -particle scattering experiment, most of the $\alpha$ -particles passed straight through the gold foil with negligible deflection, while a very small fraction was scattered through angles greater than $90^\circ$ . Which fundamental deduction about atomic structure was directly established by these rare, large-angle scatterings?

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Electrons revolve around the positive center in discrete, non-radiating stationary orbits.

The positive charge is distributed uniformly throughout a sphere of atomic dimensions containing embedded electrons.

Orbiting electrons continuously radiate energy, causing them to gradually spiral inward toward the center.

Show answer & explanation

Answer: The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Answer

The positive charge and nearly all the atomic mass are concentrated in an extremely small, dense central region called the nucleus.

Large-angle deflection (

>90^\circ

) requires an immense repulsive Coulomb force, which can only occur if the entire positive charge and virtually all atomic mass are concentrated in a tiny central region (the nucleus). If positive charge were spread out over the entire atomic volume, the maximum electric field would be far too weak to turn back high-velocity alpha particles.

Step-by-Step Solution

1

Analyze the experimental observations from Rutherford's alpha-scattering experiment.

Most

\alpha

-particles pass undeflected (indicating mostly empty space), while a tiny fraction scatter at angles

>90^\circ

.

Understanding the physical cause of large-angle electrostatic deflection.

2

Relate electrostatic repulsive force to mass and charge distribution.

To turn around a fast-moving positive

\alpha

-particle (

He^{2+}

), it must experience a intense Coulomb repulsion

F = \frac{1}{4\pi\varepsilon_0}\frac{q_1 q_2}{r^2}

at very small distance

r

.

A diffuse charge distribution (Thomson model) yields weak electric fields that cannot cause wide-angle scattering.

3

Identify the structural conclusion drawn by Rutherford.

The entire positive charge and mass must reside in a massive, concentrated center (the nucleus).

Directly matches the core physical takeaway of the nuclear atomic model.

Key Concept

Rutherford Nuclear Model and Alpha Scattering Deduction

Question 63Question →

In an X-ray tube operated at a fixed accelerating potential, replacing the target anode with a metal of higher atomic number decreases the cut-off (minimum) wavelength $\lambda_{\text{min}}$ of the continuous X-ray spectrum.

True

False

Show answer & explanation

Answer: False

Answer

False

The statement is false because, by the Duane-Hunt law (

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

), the minimum cut-off wavelength depends strictly on the accelerating potential

V

applied across the tube and fundamental physical constants. It is completely independent of the target material's atomic number

Z

.

Step-by-Step Solution

1

Identify the relationship governing the minimum (cut-off) wavelength of continuous X-rays.

By the Duane-Hunt law, the maximum energy of an emitted X-ray photon equals the total kinetic energy of an accelerating electron:

E_{\text{max}} = eV = \frac{hc}{\lambda_{\text{min}}}

.

The continuous X-ray spectrum is produced via Bremsstrahlung (braking radiation) when high-speed electrons are decelerated by the electric fields of target nuclei.

2

Express the minimum wavelength

\lambda_{\text{min}}

in terms of operational variables.

Rearranging the equation yields

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

, where

h

is Planck's constant,

c

is the speed of light,

e

is the elementary charge, and

V

is the tube potential.

This formula establishes that

\lambda_{\text{min}}

depends exclusively on the accelerating voltage

V

and physical constants.

3

Analyze the effect of altering the target material's atomic number

Z

at constant voltage

V

.

Changing the atomic number

Z

shifts characteristic spectral line wavelengths (via Moseley's law) and increases overall Bremsstrahlung intensity, but leaves

\lambda_{\text{min}}

entirely unchanged.

Since

Z

does not enter the Duane-Hunt expression for

\lambda_{\text{min}}

, varying

Z

has zero effect on the shortest emitted wavelength.

Key Concept

Duane-Hunt Law and Cut-off Wavelength Independence from Target Material

Question 64Question →

An X-ray tube operates at an initial potential difference of $V_1$ . When the potential difference is increased by $20.0\text{ kV}$ , the minimum cutoff wavelength of the emitted X-rays is reduced to one-third of its initial value. What is the initial operating potential difference $V_1$ of the X-ray tube?

10.0\text{ kV}

30.0\text{ kV}

40.0\text{ kV}

5.0\text{ kV}

Show answer & explanation

Answer: $10.0\text{ kV}$

Answer

The initial operating potential difference of the tube is

10.0\text{ kV}

.

The Duane-Hunt law states that

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

. Because minimum wavelength is inversely proportional to potential difference, reducing the minimum wavelength to one-third requires the operating potential to increase by a factor of three (

V_2 = 3V_1

). Given that

V_2 = V_1 + 20.0\text{ kV}

, setting

V_1 + 20.0\text{ kV} = 3V_1

gives

2V_1 = 20.0\text{ kV}

, which solves to

V_1 = 10.0\text{ kV}

.

Step-by-Step Solution

1

Relate the minimum cutoff wavelength to potential difference using the Duane-Hunt law.

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

, which implies

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

.

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

2

Set up the ratio between initial and final conditions.

Since

\lambda_2 = \frac{1}{3}\lambda_1

, the new voltage must be three times the initial voltage:

V_2 = 3V_1

.

Cutoff wavelength is inversely proportional to accelerating voltage.

3

Substitute

V_2 = V_1 + 20.0\text{ kV}

into the equation

V_2 = 3V_1

and solve for

V_1

.

V_1 + 20.0\text{ kV} = 3V_1 \implies 2V_1 = 20.0\text{ kV} \implies V_1 = 10.0\text{ kV}

.

Solving the linear algebraic equation yields the initial potential difference.

Key Concept

Duane-Hunt Law and Inverse Relationship between Minimum Wavelength and Accelerating Potential

Estimated Time:2m 0s

Question 65Question →

The energy of an electron in the first excited state of a hydrogen atom is $-3.4\text{ eV}$ . What is the minimum energy, in joules ( $\text{J}$ ), required to completely remove the electron from this state to ionize the atom? ( $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 5.44e-19

Answer

The minimum energy required to ionize the atom from its first excited state is

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

.

Ionization energy is defined as the minimum energy necessary to completely remove an electron from its bound atomic energy level to infinity (

E_{\infty} = 0\text{ eV}

). For an electron at

E = -3.4\text{ eV}

, the required energy change is

\Delta E = 0 - (-3.4\text{ eV}) = 3.4\text{ eV}

. Converting this to joules using

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

gives

3.4 \times 1.6 \times 10^{-19} = 5.44 \times 10^{-19}\text{ J}

.

Step-by-Step Solution

1

Determine the energy required for ionization in electron-volts (eV)

\Delta E = E_{\infty} - E_2 = 0\text{ eV} - (-3.4\text{ eV}) = 3.4\text{ eV}

Ionization requires supplying sufficient energy to raise the electron from its bound energy state (

E_2 = -3.4\text{ eV}

) to the ionization limit where it is free (

E_{\infty} = 0\text{ eV}

).

2

Convert the ionization energy from electron-volts to joules

E = 3.4\text{ eV} \times 1.6 \times 10^{-19}\text{ J/eV} = 5.44 \times 10^{-19}\text{ J}

To convert energy from electron-volts (eV) to joules (J), multiply the value in eV by the elementary charge conversion factor

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

.

Key Concept

Ionization Energy and Energy Level Transitions

Estimated Time:1m 30s

Question 66Question →

A beam of cathode rays (electrons) traveling at a speed of $4.00 \times 10^{6}\text{ m s}^{-1}$ enters a region with a uniform magnetic field of $5.00 \times 10^{-4}\text{ T}$ directed perpendicular to the beam. Taking the specific charge of an electron ( $\frac{e}{m}$ ) to be $1.60 \times 10^{11}\text{ C kg}^{-1}$ , what is the radius of the circular path traced by the cathode rays, in centimeters?

Show answer & explanation

Answer: 5

Answer

The radius of the circular path followed by the cathode rays is 5.0 cm.

When cathode rays enter a uniform magnetic field at right angles, the magnetic force acts as a centripetal force causing the electron beam to trace a circular arc of radius

r = \frac{v}{(e/m)B}

. Substituting

v = 4.00 \times 10^6\text{ m s}^{-1}

,

e/m = 1.60 \times 10^{11}\text{ C kg}^{-1}

, and

B = 5.00 \times 10^{-4}\text{ T}

yields

r = 0.05\text{ m}

, which equals

5.0\text{ cm}

.

Step-by-Step Solution

1

Set the magnetic Lorentz force equal to the required centripetal force for circular motion.

e v B = \frac{m v^2}{r}

Cathode rays consist of moving electrons experiences a magnetic force perpendicular to both their velocity and the magnetic field.

2

Express the radius

r

in terms of speed

v

, magnetic field

B

, and specific charge

\frac{e}{m}

.

r = \frac{v}{\left(\frac{e}{m}\right) B}

Simplifying the force balance equation isolates the radius on one side.

3

Substitute the given numerical values into the expression for

r

.

r = \frac{4.00 \times 10^{6}\text{ m s}^{-1}}{\left(1.60 \times 10^{11}\text{ C kg}^{-1}\right) \times \left(5.00 \times 10^{-4}\text{ T}\right)} = 0.05\text{ m}

Calculating the denominator gives

8.00 \times 10^{7}\text{ C T kg}^{-1}

, leading to

0.05\text{ m}

.

4

Convert the radius from meters to centimeters as requested by the question.

r = 0.05\text{ m} \times 100\text{ cm m}^{-1} = 5.0\text{ cm}

1 meter equals 100 centimeters.

Key Concept

Deflection of Cathode Rays in a Magnetic Field

Estimated Time:1m 30s

Question 67Question →

A beam of cathode rays is accelerated from rest through an electric potential difference $V$ before entering a uniform magnetic field $B$ applied perpendicular to the direction of motion, causing the rays to bend into a circular arc of radius $r$ . If the accelerating potential difference is increased to $2V$ and the magnetic field intensity is increased to $2B$ , what is the new radius of curvature of the cathode ray path?

\frac{r}{\sqrt{2}}

r

\sqrt{2}r

2r

Show answer & explanation

Answer: $\frac{r}{\sqrt{2}}$

Answer

The new radius of curvature is

\frac{r}{\sqrt{2}}

.

The kinetic energy gained by an electron of mass

m

and charge

e

accelerated through potential difference

V

is

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

, giving

v = \sqrt{\frac{2eV}{m}}

. When entering a perpendicular magnetic field

B

, centripetal force gives

e v B = \frac{m v^2}{r}

, leading to

r = \frac{m v}{e B} = \frac{1}{B}\sqrt{\frac{2m V}{e}}

. Replacing

V

with

2V

and

B

with

2B

gives

r' = \frac{\sqrt{2}}{2}r = \frac{r}{\sqrt{2}}

.

Step-by-Step Solution

1

Relate electron velocity to accelerating potential difference

V

v = \sqrt{\frac{2eV}{m}}

The electrical potential energy lost equals the kinetic energy gained by the cathode ray electrons:

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

.

2

Express the radius of curvature

r

in terms of

V

and

B

r = \frac{mv}{eB} = \frac{1}{B}\sqrt{\frac{2mV}{e}}

The magnetic force

evB

provides the necessary centripetal force

\frac{mv^2}{r}

.

3

Substitute the scaled values

V' = 2V

and

B' = 2B

into the radius expression

r' = \frac{1}{2B}\sqrt{\frac{2m(2V)}{e}} = \frac{\sqrt{2}}{2}\left(\frac{1}{B}\sqrt{\frac{2mV}{e}}\right) = \frac{r}{\sqrt{2}}

Increasing

V

by a factor of 2 increases

v

by

\sqrt{2}

, while doubling

B

increases the denominator by 2.

Key Concept

Deflection of cathode rays in magnetic fields and energy conversion of accelerated charges

Question 68Question →

In an X-ray tube, non-relativistic electrons accelerated from rest hit a target anode, producing continuous X-radiation with a minimum cut-off wavelength of $\lambda_0$ . If the operating potential difference across the tube is adjusted such that the maximum momentum of the colliding electrons increases by $50\%$ , what is the new cut-off wavelength of the emitted X-rays in terms of $\lambda_0$ ?

\frac{2}{3}\lambda_0

\frac{4}{9}\lambda_0

\frac{9}{4}\lambda_0

\frac{3}{2}\lambda_0

Show answer & explanation

Answer: $\frac{4}{9}\lambda_0$

Answer

The new cut-off wavelength of the emitted X-rays is

\frac{4}{9}\lambda_0

.

According to the Duane-Hunt law, the maximum photon energy produced in continuous X-radiation equals the maximum kinetic energy of the striking electrons:

E_{\text{max}} = \frac{hc}{\lambda_{\min}} = E_k

. Expressing kinetic energy in terms of momentum yields

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

, which gives

\lambda_{\min} = \frac{2mhc}{p^2}

. Therefore,

\lambda_{\min}

is inversely proportional to

p^2

. When momentum increases by

50\%

(

p_2 = 1.5 p_1 = \frac{3}{2}p_1

),

p^2

increases by a factor of

\frac{9}{4}

. Consequently, the new cut-off wavelength becomes

\frac{4}{9}\lambda_0

.

Step-by-Step Solution

1

Relate electron momentum to electron kinetic energy

The kinetic energy

E_k

of non-relativistic electrons of mass

m

with momentum

p

is

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

.

Electrons are accelerated through potential difference

V

, gaining kinetic energy

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

.

2

Apply the Duane-Hunt law for minimum X-ray wavelength

\lambda_0 = \frac{hc}{E_k} = \frac{2mhc}{p_1^2}

.

The maximum energy of an X-ray photon corresponds to the shortest (cut-off) wavelength

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

.

3

Calculate the updated momentum and new kinetic energy ratio

New momentum

p_2 = 1.5 p_1 = \frac{3}{2} p_1

, so

p_2^2 = \frac{9}{4} p_1^2

.

An increase of

50\%

means multiplying the initial momentum by

1 + 0.5 = 1.5 = \frac{3}{2}

.

4

Determine the new cut-off wavelength

\lambda_{\text{new}}

\lambda_{\text{new}} = \frac{2mhc}{p_2^2} = \frac{2mhc}{\frac{9}{4}p_1^2} = \frac{4}{9} \left(\frac{2mhc}{p_1^2}\right) = \frac{4}{9}\lambda_0

.

Since

\lambda_{\min}

is inversely proportional to

p^2

, scaling momentum by

\frac{3}{2}

reduces the cut-off wavelength by a factor of

\left(\frac{2}{3}\right)^2 = \frac{4}{9}

.

Key Concept

Duane-Hunt Law and Electron Kinetics

Question 69Question →

Ultraviolet radiation of fixed frequency $f$ , which exceeds the threshold frequency $f_0$ of a zinc emitter plate, causes photoelectron emission. If the intensity of the incident radiation is multiplied by two without altering its frequency, what is the effect on the maximum kinetic energy of the photoelectrons and the emission current?

The maximum kinetic energy is unchanged, but the emission current is doubled.

The maximum kinetic energy is doubled, and the emission current is also doubled.

The maximum kinetic energy is doubled, while the emission current remains unchanged.

The maximum kinetic energy and emission current both remain unchanged.

Show answer & explanation

Answer: The maximum kinetic energy is unchanged, but the emission current is doubled.

Answer

The maximum kinetic energy is unchanged, but the emission current is doubled.

In the photoelectric effect, the maximum kinetic energy of emitted electrons (

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

) depends only on the frequency of the incident radiation and the work function of the metal. Changing light intensity changes only the number of photons arriving per second, which doubles the rate of photoelectron emission and hence doubles the photoelectric current while leaving maximum kinetic energy unchanged.

Step-by-Step Solution

1

Analyze the dependence of maximum kinetic energy on radiation parameters using Einstein's photoelectric equation.

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

. Since frequency

f

and work function

W_0

remain constant,

K_{\text{max}}

remains unchanged.

Individual photon energy is determined strictly by frequency (

E = hf

). Intensity does not alter individual photon energy.

2

Analyze the relationship between light intensity and photoelectric current.

Intensity

I

is directly proportional to the number of incident photons per second (

N

). Since one photon liberates one electron, doubling intensity doubles the photoelectron emission rate, thereby doubling the photoelectric current.

Photoelectric current measures the rate of charge emission, which depends directly on photon flux.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 70Question →

In Bohr's model of the hydrogen atom, the energy of an electron in a stationary orbit with principal quantum number $n$ is given by $E_n = -\frac{13.6}{n^2}\text{ eV}$ . What is the frequency of the photon emitted when an electron transitions from the $n = 4$ energy state to the $n = 2$ energy state? ( $h = 6.63 \times 10^{-34}\text{ J}\cdot\text{s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

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

3.85 \times 10^{33}\text{ Hz}

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

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

Show answer & explanation

Answer: $6.15 \times 10^{14}\text{ Hz}$

Answer

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

The energy of the emitted photon corresponds to the difference between the initial energy level (

n = 4

) and the final energy level (

n = 2

), giving

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

. Converting

2.55\text{ eV}

to Joules yields

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

. Dividing this energy by Planck's constant (

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

) produces a photon frequency of

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

.

Step-by-Step Solution

1

Calculate the energy levels for

n = 4

and

n = 2

E_4 = -\frac{13.6}{4^2} = -0.85\text{ eV}

and

E_2 = -\frac{13.6}{2^2} = -3.40\text{ eV}

Electron energy in Bohr's model depends inversely on the square of the principal quantum number.

2

Find the energy of the emitted photon

\Delta E = E_4 - E_2 = -0.85\text{ eV} - (-3.40\text{ eV}) = 2.55\text{ eV}

The energy of the emitted photon equals the energy lost during the downward transition between states.

3

Convert the photon energy from electron-volts to Joules

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

Energy must be converted to SI units (Joules) before applying Planck's equation.

4

Calculate the photon frequency using

f = \frac{\Delta E}{h}

f = \frac{4.08 \times 10^{-19}\text{ J}}{6.63 \times 10^{-34}\text{ J}\cdot\text{s}} \approx 6.15 \times 10^{14}\text{ Hz}

Photon energy and frequency are related by the Planck relation

E = h f

.

Key Concept

Bohr Model Energy Transitions and Photon Frequency

Question 71Question →

A radioactive isotope has a half-life of $10\text{ hours}$ . If a sample initially contains $64\text{ g}$ of the isotope, what mass of the isotope has decayed after an elapsed time of $30\text{ hours}$ ?

8\text{ g}

16\text{ g}

48\text{ g}

56\text{ g}

Show answer & explanation

Answer: $56\text{ g}$

Answer

The mass of the isotope that has decayed after

30\text{ hours}

is

56\text{ g}

.

The correct answer is

56\text{ g}

. With a half-life of

10\text{ hours}

, an elapsed time of

30\text{ hours}

represents

3

half-lives. After

3

half-lives, the mass remaining undecayed is

64\text{ g} \times (1/2)^3 = 8\text{ g}

. Therefore, the mass that has decayed is

64\text{ g} - 8\text{ g} = 56\text{ g}

.

Step-by-Step Solution

1

Calculate the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{30\text{ hours}}{10\text{ hours}} = 3\text{ half-lives}

Determining how many half-life intervals occur during the total elapsed time.

2

Calculate the mass of the sample remaining undecayed (

N

).

N = N_0 \left(\frac{1}{2}\right)^n = 64\text{ g} \times \left(\frac{1}{2}\right)^3 = 64\text{ g} \times \frac{1}{8} = 8\text{ g}

Using the radioactive decay formula to find the remaining undecayed mass.

3

Calculate the mass of the sample that has decayed (

N_{\text{decayed}}

).

N_{\text{decayed}} = N_0 - N = 64\text{ g} - 8\text{ g} = 56\text{ g}

Subtracting the remaining mass from the initial mass to find the amount decayed.

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:45s

Question 72Question →

An electron inside an excited gas atom undergoes a transition from an upper energy level of $-2.40\text{ eV}$ to a lower energy level of $-5.15\text{ eV}$ . What is the wavelength, in nanometers ( $\text{nm}$ ), of the emitted photon? (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: 450

Answer

The wavelength of the emitted photon is 450 nm.

When an electron transitions from a higher energy level to a lower energy level, a photon is emitted with energy equal to the difference between the two energy states. Converting 2.75 eV to 4.40 x 10^-19 J and applying lambda = hc / E yields a wavelength of 4.50 x 10^-7 m, which equals 450 nm.

Step-by-Step Solution

1

Calculate the energy change of the transition in eV

\Delta E = -2.40\text{ eV} - (-5.15\text{ eV}) = 2.75\text{ eV}

The energy of the emitted photon equals the difference between the higher and lower electronic energy states.

2

Convert the transition energy into SI units (Joules)

\Delta E = 2.75 \times 1.6 \times 10^{-19}\text{ J} = 4.40 \times 10^{-19}\text{ J}

Standard physics constants h and c require energy to be expressed in Joules.

3

Calculate photon wavelength and convert to nanometers

\lambda = \frac{hc}{\Delta E} = \frac{1.98 \times 10^{-25}\text{ J m}}{4.40 \times 10^{-19}\text{ J}} = 4.50 \times 10^{-7}\text{ m} = 450\text{ nm}

Applying the de Broglie/Einstein relation connecting photon energy and wavelength.

Key Concept

Energy Level Transitions and Atomic Emission Spectra

Question 73Question →

A clean metal surface with threshold frequency $f_0$ is illuminated by monochromatic radiation of frequency $2f_0$ and light intensity $I$ , causing emission of photoelectrons with maximum kinetic energy $K_1$ and stopping potential $V_1$ . If the source is changed to radiate light of frequency $3f_0$ while its intensity is simultaneously doubled to $2I$ , what are the new maximum kinetic energy $K_2$ and stopping potential $V_2$ in terms of $K_1$ and $V_1$ ?

K_2 = 2K_1

and

V_2 = 2V_1

K_2 = 4K_1

and

V_2 = 4V_1

K_2 = 4K_1

and

V_2 = 2V_1

K_2 = 3K_1

and

V_2 = 3V_1

Show answer & explanation

Answer: $K_2 = 2K_1$ and $V_2 = 2V_1$

Answer

K_2 = 2K_1

and

V_2 = 2V_1

By Einstein's photoelectric law,

K_{\max} = hf - W_0

. For incident frequency

2f_0

and work function

W_0 = hf_0

, the initial kinetic energy is

K_1 = 2hf_0 - hf_0 = hf_0

. For incident frequency

3f_0

, the new kinetic energy is

K_2 = 3hf_0 - hf_0 = 2hf_0 = 2K_1

. Since stopping potential is related by

eV_s = K_{\max}

,

V_2 = 2V_1

. Doubling the light intensity from

I

to

2I

doubles the rate of photoelectron emission but does not alter maximum kinetic energy or stopping potential.

Step-by-Step Solution

1

Apply Einstein's photoelectric equation to the initial condition.

K_1 = h(2f_0) - hf_0 = hf_0

, and

e V_1 = K_1 = hf_0

.

The maximum kinetic energy is the energy of the incident photon minus the work function of the metal plate.

2

Apply Einstein's photoelectric equation to the new frequency condition.

K_2 = h(3f_0) - hf_0 = 2hf_0 = 2K_1

, and

e V_2 = K_2 = 2hf_0 = 2e V_1 \implies V_2 = 2V_1

.

The new photon energy is

3hf_0

, leaving

2hf_0

of excess energy as photoelectron maximum kinetic energy.

3

Evaluate the effect of doubling light intensity to

2I

.

Light intensity has zero impact on individual photoelectron kinetic energy or stopping potential; it only increases the emission rate (saturation photocurrent).

Photoelectric emission is a one-to-one photon-electron interaction process where photon frequency dictates individual energy.

Key Concept

Independence of photoelectron kinetic energy and stopping potential from light intensity

Question 74Question →

In a cathode-ray tube, electrons of mass $9.10 \times 10^{-31}\text{ kg}$ and elementary charge $1.60 \times 10^{-19}\text{ C}$ are accelerated from rest by the electric field between the cathode and the anode. What potential difference, in volts, is required to accelerate these electrons to a speed of $8.00 \times 10^{6}\text{ m s}^{-1}$ ?

Show answer & explanation

Answer: 182

Answer

The potential difference required to accelerate the electrons to the specified speed is

182\text{ V}

.

The work done by an accelerating potential difference

V

on an electron of charge

e

is converted entirely into kinetic energy

\frac{1}{2} m v^2

. Solving

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

for

V

yields

V = \frac{m v^2}{2 e} = 182\text{ V}

.

Step-by-Step Solution

1

Equate the work done by the electric field to the kinetic energy gained by an electron.

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

Work done on a charged particle moving through an electric potential difference equals its gain in kinetic energy.

2

Isolate the potential difference

V

on one side of the equation.

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

Algebraic rearrangement to solve for the target variable.

3

Substitute the given numerical parameters into the equation.

V = \frac{(9.10 \times 10^{-31}\text{ kg}) \times (8.00 \times 10^{6}\text{ m s}^{-1})^2}{2 \times (1.60 \times 10^{-19}\text{ C})}

Populating the formula with the specified values for electron mass, speed, and charge.

4

Perform the final calculation.

V = 182\text{ V}

Simplifying the numerical expression gives

182\text{ V}

.

Key Concept

Acceleration of charged particles in electric fields

Question 75Question →

An X-ray tube operates at an accelerating potential difference of $20\text{ kV}$ . What is the maximum energy of the produced X-ray photons in Joules? (Take elementary charge $e = 1.6 \times 10^{-19}\text{ C}$ )

3.2 \times 10^{-15}\text{ J}

3.2 \times 10^{-18}\text{ J}

1.25 \times 10^{23}\text{ J}

2.0 \times 10^{4}\text{ J}

Show answer & explanation

Answer: $3.2 \times 10^{-15}\text{ J}$

Answer

3.2 \times 10^{-15}\text{ J}

According to the Duane-Hunt law, the maximum energy of an emitted X-ray photon equals the maximum kinetic energy gained by an electron accelerated through potential difference

V

, which is

E_{\text{max}} = e V

. Substituting

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

and

V = 20,000\text{ V}

yields

3.2 \times 10^{-15}\text{ J}

.

Step-by-Step Solution

1

Convert the accelerating potential difference from kilovolts (kV) to volts (V).

V = 20\text{ kV} = 20,000\text{ V} = 2.0 \times 10^{4}\text{ V}

Standard SI units require potential difference in volts.

2

Apply the Duane-Hunt relationship for maximum photon energy

E_{\text{max}} = e V

.

E_{\text{max}} = 1.6 \times 10^{-19}\text{ C} \times 2.0 \times 10^{4}\text{ V} = 3.2 \times 10^{-15}\text{ J}

The kinetic energy acquired by accelerated electrons is completely converted into the maximum energy of an emitted X-ray photon.

Key Concept

Duane-Hunt Law and Maximum X-ray Photon Energy

Estimated Time:45s

Question 76Question →

Monochromatic radiation carrying photons of energy $4.8\text{ eV}$ illuminates a cesium surface inside a photoelectric cell. If the work function of cesium is $2.1\text{ eV}$ , determine the stopping potential, in volts, needed to reduce the photoelectric current to zero.

Show answer & explanation

Answer: 2.7

Answer

The stopping potential required to reduce the photoelectric current to zero is

2.7\text{ V}

.

Einstein's photoelectric equation states that incident photon energy

E

equals the work function

W_0

plus the maximum kinetic energy

K_{\text{max}}

of the photoelectrons (

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

). Rearranging gives

K_{\text{max}} = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

. Since

K_{\text{max}} = e V_s

, an electron-volt value of kinetic energy numerically equals the stopping potential in volts, giving a stopping potential of

2.7\text{ V}

.

Step-by-Step Solution

1

Calculate the maximum kinetic energy (

K_{\text{max}}

) of the emitted photoelectrons.

K_{\text{max}} = E - W_0 = 4.8\text{ eV} - 2.1\text{ eV} = 2.7\text{ eV}

According to Einstein's photoelectric equation, incident photon energy is divided into overcoming the work function of the metal and providing kinetic energy to the liberated electron.

2

Determine the stopping potential (

V_s

) from the maximum kinetic energy.

V_s = \frac{K_{\text{max}}}{e} = \frac{2.7\text{ eV}}{e} = 2.7\text{ V}

The stopping potential

V_s

is the opposing potential difference needed to stop the fastest moving photoelectrons, defined by

K_{\text{max}} = e V_s

.

Key Concept

Photoelectric Effect and Work Function

Question 77Question →

A proton of mass $1.67 \times 10^{-27}\text{ kg}$ and an electron of mass $9.11 \times 10^{-31}\text{ kg}$ , both carrying charges of equal magnitude, are accelerated from rest through the same electric potential difference. Calculate the ratio of the de Broglie wavelength of the electron to that of the proton.

Show answer & explanation

Answer: 42.8

Answer

The ratio of the de Broglie wavelength of the electron to that of the proton is 42.8.

The de Broglie wavelength of a particle accelerated through potential difference

V

is given by

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

. Since both particles carry equal charge

q

and experience the same potential

V

, the ratio of their wavelengths simplifies to

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}} = \sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

.

Step-by-Step Solution

1

Relate kinetic energy to accelerating potential difference

E_k = qV

Electric potential energy converted into kinetic energy during acceleration from rest.

2

Express de Broglie wavelength in terms of particle mass, charge, and potential difference

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

Combining

p = \sqrt{2mE_k}

with de Broglie's formula

\lambda = \frac{h}{p}

.

3

Formulate the wavelength ratio of electron to proton

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}

Planck's constant

h

, elementary charge

q

, and potential difference

V

are identical for both particles and cancel out.

4

Substitute numerical values and compute final ratio

\sqrt{\frac{1.67 \times 10^{-27}}{9.11 \times 10^{-31}}} \approx 42.8

Square root of the proton-to-electron mass ratio yields the inverse ratio of their wavelengths.

Key Concept

De Broglie wavelength relation to particle mass under constant accelerating potential

Question 78Question →

A radioactive mixture initially contains two radioisotopes, $P$ and $Q$ , such that the initial number of undecayed nuclei of $P$ is $8$ times that of $Q$ . If the half-life of isotope $P$ is $2\text{ hours}$ and the half-life of isotope $Q$ is $6\text{ hours}$ , calculate the time, in hours, after which the number of undecayed nuclei of both isotopes will be equal.

Show answer & explanation

Answer: 9

Answer

The time after which the number of undecayed nuclei of both isotopes will be equal is 9 hours.

By applying the radioactive decay law

N(t) = N_0(1/2)^{t/T_{1/2}}

to both isotopes with initial ratio

N_{P0} = 8N_{Q0}

and equating

N_P(t) = N_Q(t)

, we obtain

8 = 2^{t/3}

, which gives

t = 9\text{ hours}

.

Step-by-Step Solution

1

Write the decay equations for isotopes P and Q based on their half-lives

N_P(t) = N_{P0}\left(\frac{1}{2}\right)^{t/2}

and

N_Q(t) = N_{Q0}\left(\frac{1}{2}\right)^{t/6}

Radioactive decay follows the exponential relationship

N(t) = N_0 \left(\frac{1}{2}\right)^{t/T_{1/2}}

.

2

Apply the initial condition

N_{P0} = 8 N_{Q0}

and set the two expressions equal

8 N_{Q0} \left(\frac{1}{2}\right)^{t/2} = N_{Q0} \left(\frac{1}{2}\right)^{t/6}

The problem asks for the time

t

when both isotopes have equal remaining undecayed nuclei.

3

Divide both sides by

N_{Q0} \left(\frac{1}{2}\right)^{t/2}

and simplify the exponents

8 = \frac{(1/2)^{t/6}}{(1/2)^{t/2}} = \left(\frac{1}{2}\right)^{-t/3} = 2^{t/3}

Applying exponent laws simplifies the ratio of powers of one-half into a single base-two exponent.

4

Solve for time

t

using powers of 2

2^3 = 2^{t/3} \implies \frac{t}{3} = 3 \implies t = 9\text{ hours}

Equating the exponents of identical base 2 gives the exact time.

Key Concept

Radioactive Decay Law and Half-life for Isotope Mixtures

Estimated Time:2m 0s

Question 79Question →

A sample of a radioactive nuclide with a half-life of $6\text{ hours}$ initially has a mass of $200\text{ mg}$ . What mass of the nuclide has decayed after an elapsed time of $24\text{ hours}$ ?

187.5\text{ mg}

12.5\text{ mg}

50.0\text{ mg}

150.0\text{ mg}

Show answer & explanation

Answer: $187.5\text{ mg}$

Answer

The mass of the nuclide that has decayed after

24\text{ hours}

is

187.5\text{ mg}

.

After

24\text{ hours}

, exactly

4

half-lives (

24 / 6 = 4

) have passed. The remaining radioactive mass is

200\text{ mg} / 2^4 = 12.5\text{ mg}

. Subtracting the remaining mass from the initial mass (

200\text{ mg} - 12.5\text{ mg}

) yields

187.5\text{ mg}

as the decayed mass.

Step-by-Step Solution

1

Determine the number of half-lives (

n

) that have elapsed.

n = \frac{t}{T_{1/2}} = \frac{24\text{ hours}}{6\text{ hours}} = 4\text{ half-lives}

The total elapsed time divided by the half-life duration gives the total number of decay cycles.

2

Calculate the remaining mass (

N

) after

4

half-lives.

N = N_0 \left(\frac{1}{2}\right)^n = 200\text{ mg} \times \left(\frac{1}{2}\right)^4 = \frac{200\text{ mg}}{16} = 12.5\text{ mg}

The radioactive decay law states that after

n

half-lives, the initial quantity reduces by a factor of

2^n

.

3

Calculate the mass of the nuclide that has decayed (

N_{\text{decayed}}

).

N_{\text{decayed}} = N_0 - N = 200\text{ mg} - 12.5\text{ mg} = 187.5\text{ mg}

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

Key Concept

Radioactive Decay Law and Half-life Calculation

Question 80Question →

Consider the nuclear fusion reaction below:

\text{^{2}_{1}H} + \text{^{3}_{1}H} \rightarrow \text{^{4}_{2}He} + \text{^{1}_{0}n} + Q

Given the rest masses:
Mass of

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

Mass of

\text{^{3}_{1}H} = 3.0160\text{ u}

Mass of

\text{^{4}_{2}He} = 4.0015\text{ u}

Mass of

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

If

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

, what is the total energy released (

Q

) in this fusion reaction?

18.53\text{ MeV}

957.69\text{ MeV}

2.96 \times 10^{-12}\text{ MeV}

18.5\text{ MeV}

Show answer & explanation

Answer: $18.53\text{ MeV}$

Answer

18.53\text{ MeV}

The total initial mass of the reactants is

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

. The total final mass of the products is

4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

. Subtracting the product mass from the reactant mass gives a mass defect of

\Delta m = 0.0199\text{ u}

. Multiplying this mass defect by the conversion factor

931\text{ MeV/u}

yields an energy release of

18.53\text{ MeV}

.

Step-by-Step Solution

1

Calculate the total mass of the reactants

\text{Mass of reactants} = 2.0141\text{ u} + 3.0160\text{ u} = 5.0301\text{ u}

The total initial mass must be determined from the sum of the masses of deuterium and tritium.

2

Calculate the total mass of the products

\text{Mass of products} = 4.0015\text{ u} + 1.0087\text{ u} = 5.0102\text{ u}

The total final mass includes both the helium nucleus and the released neutron.

3

Determine the mass defect (

\Delta m

)

\Delta m = 5.0301\text{ u} - 5.0102\text{ u} = 0.0199\text{ u}

Mass defect is the difference between the total initial mass and the total final mass.

4

Convert the mass defect into energy released (

Q

)

Q = 0.0199 \times 931\text{ MeV} = 18.5269\text{ MeV} \approx 18.53\text{ MeV}

Using the mass-energy equivalence factor

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

yields the energy in

\text{MeV}

.

Key Concept

Calculation of energy released in nuclear fusion reactions using mass defect

Estimated Time:2m 0s

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