Physical Chemistry Questions and Answers – Quantum Theory

This set of Physical Chemistry Multiple Choice Questions & Answers (MCQs) focuses on “Quantum Theory”.

1. What did experiments show as properties of light?
a) It has properties of waves
b) It has properties of particles
c) It has properties of waves and particles
d) It is essentially particles that act as waves in certain scenarios
View Answer

Answer: c
Explanation: Experiments showing the interference of light are well explained by the wave theory, whereas Einstein’s photoelectric effect reveals particle like properties. Scientists were skeptical about the particle nature since photons do not have any mass, however all kinds of experimental evidence suggested particle-like nature.

2. A particle of mass m moves with velocity v and De Broglie wavelength λ. Suddenly, the particle is accelerated to 2v by a kick. What is its new De Broglie wavelength?
a) Increases by a factor of 2
b) Decreases by a factor of 2
c) Decreases by a factor of 4
d) Increases by a factor of 4
View Answer

Answer: b
Explanation: De Broglie wavelength is given by the relationship λDB = \(\frac{h}{p}\) = \(\frac{h}{mv}\), where ‘h’ is Plank’s constant, ‘p’ is particle momentum, ‘m’ is particle mass, and ‘v’ is particle velocity. Since De Broglie wavelength is inversely proportional to velocity, increasing the velocity by a factor of two decreases the De Broglie wavelength by a factor of two.

3. What is the kinetic energy (which equals to \(\frac{1}{2}\)mv2) of the ejected electrons in the photoelectric effect? (‘h’ is Plank’s constant, Φ is the work function of the metal, and ‘v’ is the light frequency)
a) hv – Φ
b) Φ – hv
c) Φ2hv
d) Φ
View Answer

Answer: a
Explanation: The energy brought in from light is hv, and the internal work function of the metal (Φ) removes some energy from the incoming light to eject the electron. The remaining energy is given to the electrons in the form of kinetic energy.

4. What is the resultant velocity of the electrons ejected from a metal when it is struck by light in the photoelectric effect?
a) \(\frac{2}{m}(hv-\phi)\)
b) \(\frac{m}{2}(hv-\phi)\)
c) \(\frac{2}{m}(\phi)\)
d) \(\sqrt{\frac{2}{m}(hv-\phi)}\)
View Answer

Answer: d
Explanation: Equating kinetic energy, light energy and work function gives \(\frac{1}{2}mv^2 = hv-\phi\). Solving this and rearranging gives \(v^2 = \frac{2}{m}(hv-\phi)\) . Taking the square root gives final electron velocity \(v = \sqrt{\frac{2}{m}(hv-\phi)}\)

5. For a piece of silver metal, the velocity of electrons emitted is 7.42 × 105 ms-1 using photons of wavelength 200nm. What is the work function of this metal in electron volts?
a) 4.7 eV
b) 6.2 eV
c) 1.5 eV
d) 10 eV
View Answer

Answer: a
Explanation: Re-arranging the mathematical expression for work function gives Φ = hv – \(\frac{1}{2}\)mv2 = h\(\frac{c}{\lambda}\) – \(\frac{1}{2}\)mv2. Plugging in numbers gives Φ = \(\frac{1239.8eV nm}{200 nm}-\frac{1}{2}\) × 0.5 × 106 eV × \((\frac{7.42×10^5}{3×10^8})\)2 = 6.2 – 1.5 = 4.7 eV

6. If an electron is accelerated through a circuit with potential difference V, what is the total energy E of the electron in classical mechanics?
a) E = V
b) E = V + \(\frac{1}{2}\)mv2
c) E = \(\frac{1}{2}\)mv2
d) E = V2
View Answer

Answer: b
Explanation: Total energy is the sum of kinetic and potential energy, each given by \(\frac{1}{2}\)mv2 and V for this electron. Summing these two gives the total energy. This expression is correct in classical mechanics, whereas in quantum mechanics one needs to change the kinetic energy term to involve quantum mechanical operators.

7. If an electron is accelerated through a circuit with potential difference V, what is the total energy E in terms of the momentum of the electron ‘p’?
a) E = V + \(\frac{p^2}{2m}\)
b) E = \(\frac{p^2}{2m}\)
c) E = V
d) E = V2
View Answer

Answer: a
Explanation: The only quantity needing replacement in the total energy expression is mass and velocity for momentum. Since p = mv, \(\frac{1}{2}\)mv2 can be replaced by \(\frac{p^2}{2m}\).

8. Classic electrodynamic theory indicates that energy density of blackbody radiation is given by ρν = \(\frac{8\pi ν^2}{c^3}\)kT. What is the main flaw of this equation that does not match experimental evidence?
a) This equation does not work well at low frequencies
b) Energy density is not directly proportional to temperature at higher temperatures
c) Experimentally determined energy density approaches zero as frequency approaches infinity (asymptotic), whereas this equation implies that energy density will go to infinity
d) This equation does not account for the spherical shape of the blackbody
View Answer

Answer: c
Explanation: Energy density can never reach infinity, and it approaches zero at very high frequencies according to experimental evidence. However, this equation works very well at lower frequencies and clearly accounts for spherical three-dimensional shape of the black body. Temperature is also found to be linear at these low frequencies.

9. The new quantum mechanical equation for energy density was determined to be ρν = \(\frac{8πh(\frac{v}{c})^3}{e^{\frac{hv}{kT}}-1}\). How does energy density change with increasing temperature?
a) It decreases inverse exponentially
b) It increases inverse exponentially
c) It increases linearly
d) It decreases linearly
View Answer

Answer: b
Explanation: Increasing temperature would decrease the exponential term in the above equation and hence make the denominator smaller. This would subsequently increase energy density inverse exponentially as the exponent is in the denominator.

10. Since the energy density decreases to zero with increasing frequency for the quantum mechanical model, there should be a maximum peak before the function decreases. Mathematically, how would one go about determining the expression for this maximum energy density in terms of frequency?
a) Partially differentiate (keeping other variables as constant) energy density with respect to frequency and equate to zero
b) Integrate the area under the function
c) Partially differentiate (keeping other variables as constant) energy density with respect to temperature and equate to zero
d) Partially differentiate (keeping other variables as constant) energy density with respect to speed of light and equate to zero
View Answer

Answer: a
Explanation: Partially differentiating in terms of frequency and equating to zero would give the maximum energy density for blackbody radiation as differentiation gives a maxima/minimum. We are only looking for a maxima in the energy density and frequency domain, hence other variables are irrelevant.

Sanfoundry Global Education & Learning Series – Physical Chemistry.

To practice all areas of Physical Chemistry, here is complete set of Multiple Choice Questions and Answers.

If you find a mistake in question / option / answer, kindly take a screenshot and email to [email protected]

Subscribe to our Newsletters (Subject-wise). Participate in the Sanfoundry Certification contest to get free Certificate of Merit. Join our social networks below and stay updated with latest contests, videos, internships and jobs!

Youtube | Telegram | LinkedIn | Instagram | Facebook | Twitter | Pinterest
Manish Bhojasia - Founder & CTO at Sanfoundry
Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

Subscribe to his free Masterclasses at Youtube & discussions at Telegram SanfoundryClasses.