Quantum Physics: Black Body Radiation, Photoelectric & Compton Effects, and Wave-Particle Duality: 40 MCQs with answers

Black body radiation

Classical theory predicted infinite energy at short wavelengths (ultraviolet catastrophe), which Planck resolved by quantizing energy.

The proportionality constant between energy and frequency

Planck's constant $h$ appears in $E = h\nu$, relating a photon's energy to its frequency.

No electrons are emitted regardless of intensity

Below threshold frequency, photon energy $h\nu$ is less than the work function $\phi$, so no emission occurs.

$K_{max} = h\nu - \phi$

The maximum kinetic energy is the photon energy minus the work function: $K_{max} = h\nu - \phi$.

Increase in wavelength of X-rays scattered by electrons

Compton scattering shows that X-ray photons lose energy to electrons, increasing their wavelength.

The wave nature of electrons

Electron diffraction patterns confirmed de Broglie's hypothesis that matter has wave-like properties.

True

Planck's constant $h = 6.626 \times 10^{-34}$ J·s is fundamental to all quantum phenomena.

Quanta

The correct term is 'quanta', which refers to the discrete packets of energy that Planck proposed in his solution to the black body radiation problem.

Work function

The minimum energy needed to remove an electron from a metal is called the 'work function'.

Energy of a photon

The equation $E = h u$ describes the energy of a photon, where $h$ is Planck's constant and $ u$ is the frequency.

Compton effect

The phenomenon where X-rays scatter off free electrons and result in a longer wavelength is known as the 'Compton effect'.

Wave-particle duality

The Davisson–Germer experiment demonstrated wave-particle duality by showing that electrons can exhibit wave-like behavior.

Probability density

In quantum mechanics, $|\psi|^2$ represents the probability density of finding a particle at a given position and time.

Energy and time

Heisenberg's uncertainty principle describes the relationship between the uncertainties in position and momentum (and also energy and time).

Work function

In the photoelectric effect equation, $\phi$ represents the work function, which is the minimum energy required to remove an electron from a metal.

It increases

As the frequency of a photon increases, its energy also increases, according to the equation $E = h u$.

Scattering angle

In the Compton effect formula, $ heta$ represents the scattering angle of the photon after it collides with an electron.

Quanta

The correct option is 'Quanta' because Max Planck introduced the concept of energy being emitted in discrete packets called quanta to explain black body radiation.

Energy is proportional to frequency

The correct option is 'Energy is proportional to frequency' because Planck's law states that the energy of each quantum is given by the equation $E = h u$, where $h$ is Planck's constant.

Work function

The correct option is 'Work function' because it represents the minimum energy needed to remove an electron from the surface of a metal.

It increases

The correct option is 'It increases' because the Compton effect shows that the wavelength of the scattered X-ray photon is longer than that of the incident photon.

Maximum kinetic energy of emitted electrons

The correct option is 'Maximum kinetic energy of emitted electrons' because this equation describes how the energy of the photon is used to overcome the work function, with the remainder being the kinetic energy of the emitted electron.

It is the frequency below which no electrons are emitted

The correct option is 'It is the frequency below which no electrons are emitted' because the photoelectric effect only occurs when the frequency of the incident light is above this threshold.

Failure to predict black body radiation correctly

The correct option is 'Failure to predict black body radiation correctly' because classical physics predicted infinite energy emission at short wavelengths, leading to the ultraviolet catastrophe.

Angle of the scattered photon

The correct option is 'Angle of the scattered photon' because $ heta$ in the Compton formula refers to the angle at which the photon is scattered after colliding with an electron.

The wavelength related to the particle's momentum

The correct option is 'The wavelength related to the particle's momentum' because the de Broglie wavelength is given by $\lambda = h/p$, showing the wave-particle duality of matter.

The probability density of finding the particle

The correct option is 'The probability density of finding the particle' because $|\psi|^2$ gives the likelihood of locating the particle at a specific position and time.

The probability density of finding the particle

According to the Born interpretation, $|\psi|^2$ gives the probability density of the particle's position.

1.5 eV

$K_{max} = h\nu - \phi = 3.5 - 2.0 = 1.5$ eV.

2.426 pm

$\Delta\lambda = \frac{h}{m_e c}(1 - \cos 90^\circ) = 2.426 \times (1 - 0) = 2.426$ pm.

$5.27 \times 10^{-25}$ kg m/s

$\Delta p \geq \frac{h}{4\pi \Delta x} = \frac{6.626 \times 10^{-34}}{4\pi \times 10^{-10}} = 5.27 \times 10^{-25}$ kg m/s.

It must be infinite at all points

A wave function must be finite, not infinite. It must be single-valued, continuous, and normalizable.

$3.313 \times 10^{-19}$ J

$E = h\nu = (6.626 \times 10^{-34})(5 \times 10^{14}) = 3.313 \times 10^{-19}$ J.

The number of emitted electrons increases with light intensity above threshold.

Above threshold, intensity determines the number of photons and thus the number of electrons, but kinetic energy depends on frequency.

The scattering angle of the photon

$\theta$ is the angle through which the photon is scattered, not the electron.

Forgot that $\cos 0^\circ = 1$, giving zero shift

At $0^\circ$, $\cos 0^\circ = 1$, so $\Delta\lambda = \frac{h}{m_e c}(1-1) = 0$. There is no shift for forward scattering.

It is halved.

According to Wien's displacement law, the peak wavelength $\lambda_{max}$ is inversely proportional to the temperature $T$: $\lambda_{max} = rac{b}{T}$, where $b$ is Wien's displacement constant. Thus, if the temperature is doubled, the peak wavelength is halved.

3.5 eV

First, calculate the energy of the incident photons: $E = h u = (6.626 imes 10^{-34})(6 imes 10^{14}) \approx 3.976 imes 10^{-19}$ J = $2.48$ eV. The maximum kinetic energy is given by $K_{max} = h u - \phi = 2.48\ ext{eV} - 3.0\ ext{eV} = -0.52\ ext{eV}…

3.639 pm

Using the Compton shift formula: $$\Delta\lambda = rac{h}{m_e c}(1 - \cos heta)$$, where $ heta = 135^\circ$, we find $\Delta\lambda = 2.426\ ext{pm} imes (1 - \cos 135^\circ) = 2.426\ ext{pm} imes (1 - (-rac{1}{\sqrt{2}})) = 2.426\ ext{pm} imes (1 + 0.707) = 3.639\ ext{pm}$.

5.27 imes 10^{-25} kg m/s

Using the uncertainty relation $\Delta x \cdot \Delta p \geq rac{h}{4\pi}$, we find $\Delta p \geq rac{6.626 imes 10^{-34}}{4\pi (1 imes 10^{-10})} \approx 5.27 imes 10^{-25}\ ext{kg m/s}$. This is the minimum uncertainty in momentum, not position.