Atomic and Nuclear Physics Practice Questions — JAMB UTME — Page 8

Question 141Question →

In an industrial non-destructive testing setup, an X-ray tube produces continuous X-radiation with a minimum cut-off wavelength of $3.3 \times 10^{-11}\text{ m}$ . What is the operating accelerating potential difference of the tube in kilovolts ( $\text{kV}$ )? (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 electron charge $e = 1.6 \times 10^{-19}\text{ C}$ ).

Show answer & explanation

Answer: 37.5

Answer

The operating accelerating potential difference of the tube is 37.5 kV.

According to the Duane-Hunt law, the maximum kinetic energy of electrons hitting the target equal the maximum photon energy produced:

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

. Rearranging to solve for voltage gives

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

. Substituting

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

,

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

,

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

, and

\lambda_{\text{min}} = 3.3 \times 10^{-11}\text{ m}

yields

V = 37,500\text{ V}

, which equals

37.5\text{ kV}

.

Step-by-Step Solution

1

Relate the maximum electron kinetic energy to the shortest X-ray photon wavelength using Duane-Hunt law

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

At the Duane-Hunt cutoff limit, the entire kinetic energy of an accelerating electron is converted into a single X-ray photon.

2

Rearrange the equation to solve for the accelerating voltage

V

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

Isolating

V

allows direct calculation from known fundamental constants and the given minimum wavelength.

3

Substitute the physical constants and calculate the value of

V

in Volts

V = \frac{(6.6 \times 10^{-34}\text{ J s})(3.0 \times 10^8\text{ m s}^{-1})}{(1.6 \times 10^{-19}\text{ C})(3.3 \times 10^{-11}\text{ m})} = 37,500\text{ V}

Carrying out arithmetic with scientific notation powers yields

3.75 \times 10^4\text{ V}

.

4

Convert the potential difference from Volts (V) to kilovolts (kV)

37,500\text{ V} = 37.5\text{ kV}

Dividing by 1000 converts potential difference into the requested kilovolt unit.

Key Concept

Duane-Hunt Law and Cut-off Wavelength in X-ray Production

Question 142Question →

An atom has three stationary energy levels given by $E_1 = -12.50\text{ eV}$ , $E_2 = -6.80\text{ eV}$ , and $E_3 = -3.50\text{ eV}$ . What is the wavelength, in nanometers ( $\text{nm}$ ), of the photon emitted during the transition that produces the longest wavelength line in its emission spectrum? (Take Planck's constant $h = 6.6 \times 10^{-34}\text{ J}\cdot\text{s}$ , speed of light $c = 3.0 \times 10^8\text{ m/s}$ , and $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

Show answer & explanation

Answer: 375

Answer

The wavelength of the photon emitted for the longest wavelength spectral line is 375 nm.

Photon wavelength is related to transition energy by

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

. To find the longest wavelength spectral line, the transition with the smallest energy gap must be used. Evaluating all emission transitions between the levels gives

\Delta E_{3 \to 2} = 3.30\text{ eV}

,

\Delta E_{2 \to 1} = 5.70\text{ eV}

, and

\Delta E_{3 \to 1} = 9.00\text{ eV}

. The minimum energy difference is

3.30\text{ eV}

. Converting

3.30\text{ eV}

to Joules gives

3.30 \times 1.6 \times 10^{-19} = 5.28 \times 10^{-19}\text{ J}

. Substituting this into the wavelength formula yields

\lambda = \frac{6.6 \times 10^{-34} \times 3.0 \times 10^8}{5.28 \times 10^{-19}} = 3.75 \times 10^{-7}\text{ m} = 375\text{ nm}

.

Step-by-Step Solution

1

Determine which electronic transition yields the longest wavelength photon.

Transition from

E_3

to

E_2

yields the minimum energy difference of

3.30\text{ eV}

.

Since

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

, the longest wavelength corresponds to the smallest energy transition.

2

Convert the transition energy from electron-volts to Joules.

\Delta E = 5.28 \times 10^{-19}\text{ J}

.

SI units are required for calculations involving Planck's constant and the speed of light.

3

Calculate the wavelength

\lambda

using the photon energy formula

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

.

\lambda = 375\text{ nm}

.

Substituting

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

,

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

, and

\Delta E = 5.28 \times 10^{-19}\text{ J}

gives

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

, which equals

375\text{ nm}

.

Key Concept

Inverse relationship between transition energy and photon wavelength in atomic emission spectra

Question 143Question →

A metal emitter with a work function of $3.30\text{ eV}$ is illuminated by monochromatic light of frequency $1.20 \times 10^{15}\text{ Hz}$ . If the intensity of the light is doubled while keeping its frequency constant, what is the maximum kinetic energy of the emitted photoelectrons? ( $h = 6.6 \times 10^{-34}\text{ J s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

1.65\text{ eV}

3.30\text{ eV}

4.95\text{ eV}

6.60\text{ eV}

Show answer & explanation

Answer: $1.65\text{ eV}$

Answer

The maximum kinetic energy of the emitted photoelectrons is

1.65\text{ eV}

.

According to Einstein's photoelectric equation, the maximum kinetic energy of emitted photoelectrons depends exclusively on the frequency of incident radiation and the work function of the metal (

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

). Here,

hf = 4.95\text{ eV}

and

W_0 = 3.30\text{ eV}

, yielding

K_{\text{max}} = 1.65\text{ eV}

. Increasing the light intensity increases the rate of photon arrivals and hence the rate of photoelectron emission, but leaves the kinetic energy of individual electrons unchanged.

Step-by-Step Solution

1

Calculate the energy of the incident photon in Joules using

E = hf

E = (6.6 \times 10^{-34}\text{ J s}) \times (1.20 \times 10^{15}\text{ Hz}) = 7.92 \times 10^{-19}\text{ J}

The energy delivered by each quantum of light is proportional to its frequency.

2

Convert the photon energy from Joules to electron-volts (eV)

E = \frac{7.92 \times 10^{-19}\text{ J}}{1.6 \times 10^{-19}\text{ J/eV}} = 4.95\text{ eV}

Converting to electron-volts allows direct comparison with the given work function.

3

Apply Einstein's photoelectric equation:

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

K_{\text{max}} = 4.95\text{ eV} - 3.30\text{ eV} = 1.65\text{ eV}

The maximum kinetic energy equals the excess energy of the photon after overcoming the work function.

4

Evaluate the effect of doubling light intensity at constant frequency

The maximum kinetic energy remains

1.65\text{ eV}

.

Light intensity determines the number of photons per second (current), but does not alter individual photon energy or the maximum kinetic energy per photoelectron.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 144Question →

X-rays undergo diffraction when passed through crystalline solids because the interplanar atomic spacing of a crystal lattice is comparable to the wavelength of X-rays.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. X-rays have wavelengths of approximately

10^{-10}\text{ m}

, which is of the same order of magnitude as the spacing between adjacent atomic planes in crystal lattices, allowing crystals to act as natural diffraction gratings.

The statement is true because wave diffraction requires the grating aperture or obstacle spacing to be of the same order of magnitude as the incident wavelength. Since X-ray wavelengths (approx.

10^{-10}\text{ m}

) match the interatomic spacing of crystals, crystalline solids act as effective diffraction gratings.

Step-by-Step Solution

1

Identify the characteristic wavelength range of X-rays.

X-rays are high-energy electromagnetic waves with wavelengths typically ranging from

10^{-11}\text{ m}

to

10^{-8}\text{ m}

(around

0.1\text{ nm}

).

Wave diffraction requires the width of the aperture or obstacle spacing to be comparable in size to the wavelength of the incident wave.

2

Determine the interatomic spacing of crystalline solids.

The distance between neighboring atomic planes in a typical crystal lattice is on the order of

10^{-10}\text{ m}

(

0.1\text{ nm}

to

0.3\text{ nm}

).

Comparing these dimensions shows that crystal lattice spacing matches X-ray wavelengths.

3

Evaluate the condition for wave diffraction.

Because the interatomic spacing matches the X-ray wavelength, constructive and destructive interference occurs, forming diffraction patterns.

This confirms that the given statement is true.

Key Concept

X-ray Diffraction by Crystal Lattices

Estimated Time:1m 0s

Question 145Question →

In a standard Coolidge X-ray tube setup, the filament heating current is increased while maintaining a constant accelerating potential difference between the cathode and the target anode. Which of the following changes will be observed in the emitted X-ray beam?

The intensity of the X-ray beam increases while its minimum cut-off wavelength remains unchanged

The penetrating power of the X-ray beam increases while its intensity remains constant

The minimum cut-off wavelength of the X-ray beam decreases while its intensity increases

Both the maximum frequency and the overall intensity of the X-ray beam increase

Show answer & explanation

Answer: The intensity of the X-ray beam increases while its minimum cut-off wavelength remains unchanged

Answer

The intensity of the X-ray beam increases while its minimum cut-off wavelength remains unchanged

Increasing the filament current raises cathode temperature, causing thermionic emission of more electrons per second. This increases the total number of X-ray photons emitted per second, which means the intensity of the X-ray beam increases. Since the accelerating potential difference is held constant, the maximum kinetic energy imparted to each electron (

e V

) is unchanged. Consequently, the minimum cut-off wavelength given by

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

remains constant.

Step-by-Step Solution

1

Determine the physical effect of increasing the filament heating current

Higher filament current increases filament temperature, releasing a greater number of electrons per second via thermionic emission.

The rate of electron emission directly determines the number of X-ray photons generated per unit time, which defines the beam intensity.

2

Determine the physical effect of a constant accelerating potential difference

The maximum kinetic energy of striking electrons (

E_{\max} = e V

) and minimum wavelength (

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

) remain constant.

The accelerating voltage controls individual electron energy, which governs X-ray hardness and minimum cut-off wavelength via the Duane-Hunt law.

Key Concept

Intensity vs Hardness Control in X-ray Tubes

Estimated Time:1m 0s

Question 146Question →

A photosensitive metal plate with a work function of $2.3\text{ eV}$ is illuminated by light of frequency $8.0 \times 10^{14}\text{ Hz}$ . If the intensity of the light source is doubled while maintaining the same frequency, what will be the new maximum kinetic energy of the emitted photoelectrons? ( $h = 6.6 \times 10^{-34}\text{ J s}$ , $1\text{ eV} = 1.6 \times 10^{-19}\text{ J}$ )

remain unchanged at

1.0\text{ eV}

double to

2.0\text{ eV}

quadruple to

4.0\text{ eV}

increase to

3.3\text{ eV}

Show answer & explanation

Answer: remain unchanged at $1.0\text{ eV}$

Answer

The maximum kinetic energy will remain unchanged at

1.0\text{ eV}

.

According to Einstein's photoelectric equation,

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

. The maximum kinetic energy of an emitted photoelectron depends exclusively on the frequency of the incident photons and the work function of the target metal. Increasing the light intensity increases the rate of photon arrival and thus the rate of photoelectron emission, but it does not change the energy of individual photons. Therefore, the maximum kinetic energy remains constant at

1.0\text{ eV}

.

Step-by-Step Solution

1

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

E = hf = 6.6 \times 10^{-34}\text{ J s} \times 8.0 \times 10^{14}\text{ Hz} = 5.28 \times 10^{-19}\text{ J}

. In eV:

E = \frac{5.28 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.3\text{ eV}

.

Einstein's photoelectric equation requires comparing photon energy with the work function of the metal.

2

Calculate the maximum kinetic energy of the photoelectrons using Einstein's photoelectric equation

K_{\text{max}} = E - W_0 = 3.3\text{ eV} - 2.3\text{ eV} = 1.0\text{ eV}

.

The maximum kinetic energy is the surplus energy after overcoming the metal's work function.

3

Analyze the effect of doubling light intensity at constant frequency

Doubling intensity increases the photon flux (number of photons per second), thereby increasing emission current, but leaves individual photon energy and

K_{\text{max}}

completely unchanged at

1.0\text{ eV}

.

Kinetic energy of individual photoelectrons depends strictly on photon frequency, not beam intensity.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Question 147Question →

In a photoelectric effect experiment using light of a frequency greater than the threshold frequency of a metal surface, doubling the intensity of the incident light doubles the maximum kinetic energy of the emitted photoelectrons.

True

False

Show answer & explanation

Answer: False

Answer

The statement is false. Doubling the light intensity increases the number of emitted photoelectrons per second but leaves their maximum kinetic energy unchanged.

The statement is false because the maximum kinetic energy of emitted photoelectrons is governed by Einstein's equation

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

. It depends strictly on the frequency

f

of incident light and the metal work function

W_0

. Doubling the light intensity increases the rate of photon bombardment, which increases the number of photoelectrons emitted per unit time (photoelectric current), but leaves the maximum kinetic energy of individual photoelectrons completely unchanged.

Step-by-Step Solution

1

Identify the factors determining individual photon energy and maximum kinetic energy

Individual photon energy is given by

E = hf

, and maximum photoelectron kinetic energy is

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

.

Einstein's photoelectric equation governs the energy exchange between a single incident photon and a single bound electron.

2

Analyze the physical meaning of light intensity in quantum terms

Intensity

I

is proportional to the number of photons striking the surface per unit time, not the energy of individual photons.

At a constant frequency

f

, changing intensity varies photon flux while individual photon energy

hf

stays constant.

3

Evaluate the effect of doubling light intensity on maximum kinetic energy

Doubling intensity doubles the rate of photoemission (photoelectric current) but does not change

K_{\text{max}}

.

Because

f

and

W_0

remain constant,

K_{\text{max}}

remains strictly unchanged.

Key Concept

Independence of photoelectron kinetic energy from light intensity

Estimated Time:1m 0s

Question 148Question →

A beam of monochromatic light strikes a photosensitive metal surface, causing the emission of photoelectrons. If the intensity of the incident light is doubled while maintaining the same frequency, what happens to the stopping potential of the emitted electrons?

It remains unchanged

It is doubled

It is quadrupled

It is reduced by half

Show answer & explanation

Answer: It remains unchanged

Answer

The stopping potential remains unchanged because it depends only on the frequency of the incident light, not its intensity.

According to Einstein's photoelectric theory, the maximum kinetic energy of emitted electrons is given by

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

, where

h

is Planck's constant,

f

is the frequency of the incident light, and

W_0

is the work function of the metal. The stopping potential

V_s

is related to

K_{\text{max}}

by

e V_s = K_{\text{max}}

. Doubling the intensity of the light increases the number of photons hitting the surface per unit time (thus increasing the photocurrent), but it does not change the energy of individual photons (

hf

). Consequently, the maximum kinetic energy and the stopping potential remain completely unchanged.

Step-by-Step Solution

1

Recall Einstein's photoelectric equation

E_k^{\text{max}} = hf - W_0

The maximum kinetic energy of emitted photoelectrons depends directly on the photon energy (

hf

) and the work function (

W_0

) of the metal.

2

Relate maximum kinetic energy to stopping potential

e V_s = E_k^{\text{max}} = hf - W_0 \implies V_s = \frac{hf - W_0}{e}

Stopping potential (

V_s

) is directly proportional to the maximum kinetic energy.

3

Analyze the effect of changing light intensity at constant frequency

Increasing intensity increases the number of photons per second (and thus the photocurrent), but does not alter the energy of individual photons (

hf

). Therefore,

V_s

remains constant.

Intensity affects electron rate emission, not individual electron energy.

Key Concept

Independence of photoelectron kinetic energy and stopping potential from light intensity

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