INTRODUCTION TO QUANTUM
MECHANICS COMPLETE EXAM
150 CHALLENGING QUESTIONS
WITH BOLD ANSWERS AND
DETAILED ITALICIZED
RATIONALES PREMIUM STUDY
GUIDE 2026 2027
Introduction to Quantum Mechanics Exam
1. An experimentalist shines monochromatic ultraviolet light
with a wavelength of 250 nm onto a clean potassium surface
(work function \(\Phi = 2.30\text{ eV}\)). Due to an
experimental flaw, a retarding potential of 1.50 V is applied
across the apparatus. What is the fate of the emitted
photoelectrons?
A) No electrons reach the collector because the stopping
potential is exactly 2.66 V.
B) Electrons are emitted with a maximum kinetic energy of
4.96 eV and easily reach the collector.
C) The maximum kinetic energy of the emitted electrons is
2.66 eV; thus, the fastest electrons reach the collector with a
, residual kinetic energy of 1.16 eV.
D) No electrons are emitted because the photon energy is
strictly less than the work function of potassium.
CORRECT ANSWER: C
RATiONAlE: First, find the energy of the incident photons
using \(E = hc/\lambda\). Here, \(E = (1240\text{
eV}\cdot\text{nm}) / 250\text{ nm} = 4.96\text{ eV}\).
According to Einstein's photoelectric equation, the maximum
kinetic energy of the emitted photoelectrons is \(K_{\max} = E -
\Phi = 4.96\text{ eV} - 2.30\text{ eV} = 2.66\text{ eV}\). Since
the retarding potential is 1.50 V, it opposes the motion of the
electrons, removing 1.50 eV of kinetic energy. The maximum
residual kinetic energy of the electrons reaching the collector is
\(2.66\text{ eV} - 1.50\text{ eV} = 1.16\text{ eV}\). Because this
value is greater than zero, the fastest electrons successfully reach
the collector.
2. A high-energy X-ray photon with an initial wavelength of
\(\lambda_0 = 0.0200\text{ nm}\) undergoes Compton
scattering off a stationary, unbound electron. If the photon
scatters at an angle of exactly \(90^{\circ }\) relative to its
incident direction, what is the kinetic energy transferred to
the target electron?
A) 62.0 keV
B) 6.74 keV
C) 55.3 keV
D) 11.1 keV
,CORRECT ANSWER: B
RATiONAlE: The Compton scattering formula for wavelength
shift is \(\Delta \lambda = \lambda' - \lambda_0 =
\lambda_C(1 - \cos\theta)\), where \(\lambda_C =
\frac{h}{m_e c} \approx 0.00243\text{ nm}\) is the Compton
wavelength of an electron. For \(\theta = 90^\circ\),
\(\cos(90^\circ) = 0\), so \(\Delta \lambda = \lambda_C =
0.00243\text{ nm}\). The scattered photon wavelength is
\(\lambda' = 0.0200\text{ nm} + 0.00243\text{ nm} =
0.02243\text{ nm}\). The initial energy of the photon is \(E_0 =
hc/\lambda_0 = 1240\text{ eV}\cdot\text{nm} / 0.0200\text{
nm} = 62,000\text{ eV} = 62.0\text{ keV}\). The final energy of
the photon is \(E' = hc/\lambda' = 1240\text{ eV}\cdot\text{nm}
/ 0.02243\text{ nm} = 55.28\text{ keV}\). By conservation of
energy, the kinetic energy transferred to the electron is \(K_e =
E_0 - E' = 62.0\text{ keV} - 55.28\text{ keV} \approx 6.72\text{
keV}\), which matches option B closest allowing for minor
rounding.
3. Suppose a hypothetical hydrogen-like atom consists of a
single electron orbiting a massive nucleus with a atomic
number \(Z\). If the radius of the \(n=3\) Bohr orbit is
measured to be exactly equal to the ground-state Bohr radius
of a standard hydrogen atom (\(a_{0}\)), what is the value
of \(Z\)?
A) \(Z = 3\)
B) \(Z = 6\)
C) \(Z = 9\)
D) \(Z = 27\)
, CORRECT ANSWER: C
RATiONAlE: In the Bohr model, the radius of an orbit is given
by \(r_n = \frac{n^2 a_0}{Z}\), where \(a_{0}\) is the Bohr
radius of a standard hydrogen atom (\(Z=1, n=1\)). We are
given that for the hypothetical atom at \(n=3\), the radius
\(r_{3}\) is equal to \(a_{0}\). Setting up the equation: \(a_0 =
\frac{3^2 a_0}{Z} \implies a_0 = \frac{9 a_0}{Z}\). Solving for
\(Z\) yields \(Z = 9\).
4. An electron and an alpha particle (helium nucleus,
\(m_\alpha \approx 7300 \cdot m_e\)) are accelerated from
rest through identical, non-relativistic electrical potential
differences \(V\). What is the ratio of the de Broglie
wavelength of the electron to that of the alpha particle
(\(\lambda_e / \lambda_\alpha\))?
A) \(\sqrt{7300}\)
B) \(7300\)
C) \(1 / \sqrt{7300}\)
D) \(2\sqrt{7300}\)
CORRECT ANSWER: D
RATiONAlE: The kinetic energy gained by a charged particle
accelerated through a potential difference \(V\) is \(K = qV\).
The de Broglie wavelength is \(\lambda = \frac{h}{p} =
\frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}\). For the
electron, the charge is \(e\) and mass is \(m_{e}\), giving
\(\lambda_e = \frac{h}{\sqrt{2m_e eV}}\). For the alpha
particle, the charge is \(2e\) and mass is \(m_\alpha \approx
7300 m_e\), giving \(\lambda_\alpha = \frac{h}{\sqrt{2(7300
m_e)(2e)V}} = \frac{h}{\sqrt{4 \cdot 7300 m_e eV}}\). Taking
MECHANICS COMPLETE EXAM
150 CHALLENGING QUESTIONS
WITH BOLD ANSWERS AND
DETAILED ITALICIZED
RATIONALES PREMIUM STUDY
GUIDE 2026 2027
Introduction to Quantum Mechanics Exam
1. An experimentalist shines monochromatic ultraviolet light
with a wavelength of 250 nm onto a clean potassium surface
(work function \(\Phi = 2.30\text{ eV}\)). Due to an
experimental flaw, a retarding potential of 1.50 V is applied
across the apparatus. What is the fate of the emitted
photoelectrons?
A) No electrons reach the collector because the stopping
potential is exactly 2.66 V.
B) Electrons are emitted with a maximum kinetic energy of
4.96 eV and easily reach the collector.
C) The maximum kinetic energy of the emitted electrons is
2.66 eV; thus, the fastest electrons reach the collector with a
, residual kinetic energy of 1.16 eV.
D) No electrons are emitted because the photon energy is
strictly less than the work function of potassium.
CORRECT ANSWER: C
RATiONAlE: First, find the energy of the incident photons
using \(E = hc/\lambda\). Here, \(E = (1240\text{
eV}\cdot\text{nm}) / 250\text{ nm} = 4.96\text{ eV}\).
According to Einstein's photoelectric equation, the maximum
kinetic energy of the emitted photoelectrons is \(K_{\max} = E -
\Phi = 4.96\text{ eV} - 2.30\text{ eV} = 2.66\text{ eV}\). Since
the retarding potential is 1.50 V, it opposes the motion of the
electrons, removing 1.50 eV of kinetic energy. The maximum
residual kinetic energy of the electrons reaching the collector is
\(2.66\text{ eV} - 1.50\text{ eV} = 1.16\text{ eV}\). Because this
value is greater than zero, the fastest electrons successfully reach
the collector.
2. A high-energy X-ray photon with an initial wavelength of
\(\lambda_0 = 0.0200\text{ nm}\) undergoes Compton
scattering off a stationary, unbound electron. If the photon
scatters at an angle of exactly \(90^{\circ }\) relative to its
incident direction, what is the kinetic energy transferred to
the target electron?
A) 62.0 keV
B) 6.74 keV
C) 55.3 keV
D) 11.1 keV
,CORRECT ANSWER: B
RATiONAlE: The Compton scattering formula for wavelength
shift is \(\Delta \lambda = \lambda' - \lambda_0 =
\lambda_C(1 - \cos\theta)\), where \(\lambda_C =
\frac{h}{m_e c} \approx 0.00243\text{ nm}\) is the Compton
wavelength of an electron. For \(\theta = 90^\circ\),
\(\cos(90^\circ) = 0\), so \(\Delta \lambda = \lambda_C =
0.00243\text{ nm}\). The scattered photon wavelength is
\(\lambda' = 0.0200\text{ nm} + 0.00243\text{ nm} =
0.02243\text{ nm}\). The initial energy of the photon is \(E_0 =
hc/\lambda_0 = 1240\text{ eV}\cdot\text{nm} / 0.0200\text{
nm} = 62,000\text{ eV} = 62.0\text{ keV}\). The final energy of
the photon is \(E' = hc/\lambda' = 1240\text{ eV}\cdot\text{nm}
/ 0.02243\text{ nm} = 55.28\text{ keV}\). By conservation of
energy, the kinetic energy transferred to the electron is \(K_e =
E_0 - E' = 62.0\text{ keV} - 55.28\text{ keV} \approx 6.72\text{
keV}\), which matches option B closest allowing for minor
rounding.
3. Suppose a hypothetical hydrogen-like atom consists of a
single electron orbiting a massive nucleus with a atomic
number \(Z\). If the radius of the \(n=3\) Bohr orbit is
measured to be exactly equal to the ground-state Bohr radius
of a standard hydrogen atom (\(a_{0}\)), what is the value
of \(Z\)?
A) \(Z = 3\)
B) \(Z = 6\)
C) \(Z = 9\)
D) \(Z = 27\)
, CORRECT ANSWER: C
RATiONAlE: In the Bohr model, the radius of an orbit is given
by \(r_n = \frac{n^2 a_0}{Z}\), where \(a_{0}\) is the Bohr
radius of a standard hydrogen atom (\(Z=1, n=1\)). We are
given that for the hypothetical atom at \(n=3\), the radius
\(r_{3}\) is equal to \(a_{0}\). Setting up the equation: \(a_0 =
\frac{3^2 a_0}{Z} \implies a_0 = \frac{9 a_0}{Z}\). Solving for
\(Z\) yields \(Z = 9\).
4. An electron and an alpha particle (helium nucleus,
\(m_\alpha \approx 7300 \cdot m_e\)) are accelerated from
rest through identical, non-relativistic electrical potential
differences \(V\). What is the ratio of the de Broglie
wavelength of the electron to that of the alpha particle
(\(\lambda_e / \lambda_\alpha\))?
A) \(\sqrt{7300}\)
B) \(7300\)
C) \(1 / \sqrt{7300}\)
D) \(2\sqrt{7300}\)
CORRECT ANSWER: D
RATiONAlE: The kinetic energy gained by a charged particle
accelerated through a potential difference \(V\) is \(K = qV\).
The de Broglie wavelength is \(\lambda = \frac{h}{p} =
\frac{h}{\sqrt{2mK}} = \frac{h}{\sqrt{2mqV}}\). For the
electron, the charge is \(e\) and mass is \(m_{e}\), giving
\(\lambda_e = \frac{h}{\sqrt{2m_e eV}}\). For the alpha
particle, the charge is \(2e\) and mass is \(m_\alpha \approx
7300 m_e\), giving \(\lambda_\alpha = \frac{h}{\sqrt{2(7300
m_e)(2e)V}} = \frac{h}{\sqrt{4 \cdot 7300 m_e eV}}\). Taking
