Work Function & Threshold — MH-CET Physics MCQs with Solutions
Free MH-CET Physics Work Function & Threshold MCQs with step-by-step solutions (21 questions). Part of Dual Nature of Radiation and Matter. Practise online on Prepizo — no login needed.
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Questions with solutions
Q1 — Work Function & Threshold · easy · theory
The work function of a metal is:
A. The minimum energy needed to free an electron from its surface ✓ Correct
B. The energy of the incident photon
C. The total energy of all the electrons in the metal
D. The kinetic energy of the fastest photoelectron
Solution: It represents the binding of the least tightly held electrons to the metal surface.
Q2 — Work Function & Threshold · easy · theory
The work function and threshold frequency are related by:
A. $\phi_0 = \dfrac{h}{\nu_0}$
B. $\phi_0 = h\nu_0^2$
C. $\phi_0 = \dfrac{\nu_0}{h}$
D. $\phi_0 = h\nu_0$ ✓ Correct
Solution: The threshold frequency is simply the work function expressed as a photon frequency.
Q3 — Work Function & Threshold · medium · theory
The threshold wavelength of a metal of work function $\phi_0$ is:
A. $\dfrac{\phi_0}{hc}$
B. $\dfrac{h}{\phi_0}$
C. $\dfrac{hc}{\phi_0}$ ✓ Correct
D. $hc\phi_0$
Solution: It is the longest wavelength that can still just eject an electron.
Q4 — Work Function & Threshold · medium · theory
Alkali metals such as caesium are commonly used in photocells because they have:
A. Very high densities
B. Very high work functions
C. High melting points
D. Low work functions, so visible light can eject electrons ✓ Correct
Solution: A low work function means the threshold wavelength lies in the visible range.
Q5 — Work Function & Threshold · medium · theory
The work function of a surface depends on:
A. The intensity of the incident light
B. The nature of the metal and the condition of its surface ✓ Correct
C. The stopping potential applied
D. The frequency of the incident light
Solution: It is a property of the material itself, unchanged by the light used to probe it.
Q6 — Work Function & Threshold · medium · theory
Photoemission does not occur when the incident wavelength is:
A. Equal to the threshold wavelength
B. Less than the threshold wavelength
C. In the ultraviolet region
D. Greater than the threshold wavelength ✓ Correct
Solution: A longer wavelength means a lower frequency and therefore too little energy per photon.
Q7 — Work Function & Threshold · easy · theory
The work function of a metal is usually expressed in:
A. Electron volt ✓ Correct
B. Volt
C. Coulomb
D. Newton
Solution: One electron volt is $1.6 \times 10^{-19}\text{ J}$, a convenient size for atomic-scale energies.
Q8 — Work Function & Threshold · medium · theory
The threshold wavelength of a metal is the:
A. Minimum wavelength that can cause photoemission
B. Wavelength of the emitted electrons
C. Wavelength at which the current is maximum
D. Maximum wavelength that can cause photoemission ✓ Correct
Solution: Wavelength and energy are inversely related, so the threshold sets an upper limit on wavelength.
Q9 — Work Function & Threshold · medium · numerical
The work function of a surface is $2.5\text{ eV}$. Its threshold wavelength is ($hc = 1240\text{ eV}\cdot\text{nm}$):
A. $248\text{ nm}$
B. $496\text{ nm}$ ✓ Correct
C. $620\text{ nm}$
D. $310\text{ nm}$
Solution: $\lambda_0 = \dfrac{hc}{\phi_0} = \dfrac{1240}{2.5} = 496\text{ nm}$.
Q10 — Work Function & Threshold · medium · numerical
The work function of a metal is $2\text{ eV}$. Its threshold wavelength is:
A. $1240\text{ nm}$
B. $496\text{ nm}$
C. $310\text{ nm}$
D. $620\text{ nm}$ ✓ Correct
Solution: $\lambda_0 = \dfrac{1240}{2} = 620\text{ nm}$.
Q11 — Work Function & Threshold · medium · numerical
A metal has threshold wavelength $500\text{ nm}$. Its work function is approximately:
A. $4.96\text{ eV}$
B. $0.4\text{ eV}$
C. $1.24\text{ eV}$
D. $2.48\text{ eV}$ ✓ Correct
Solution: $\phi_0 = \dfrac{1240}{500} = 2.48\text{ eV}$.
Q12 — Work Function & Threshold · medium · numerical
Photoelectric emission from a metal ceases completely when the wavelength is increased to $12000\text{ \AA}$. The threshold wavelength of the metal is:
A. $12000\text{ \AA}$ ✓ Correct
B. $3000\text{ \AA}$
C. $6000\text{ \AA}$
D. $4000\text{ \AA}$
Solution: Emission stops exactly at the threshold, so that wavelength is $\lambda_0$.
Q13 — Work Function & Threshold · medium · numerical
A metal has work function $4.13\text{ eV}$. Its threshold wavelength is approximately:
A. $400\text{ nm}$
B. $300\text{ nm}$ ✓ Correct
C. $600\text{ nm}$
D. $150\text{ nm}$
Solution: $\lambda_0 = \dfrac{1240}{4.13} \approx 300\text{ nm}$.
Q14 — Work Function & Threshold · hard · numerical
A metal has work function $3.2 \times 10^{-19}\text{ J}$. Its threshold frequency is approximately:
A. $4.83 \times 10^{15}\text{ Hz}$
B. $2.12 \times 10^{15}\text{ Hz}$
C. $4.83 \times 10^{14}\text{ Hz}$ ✓ Correct
D. $6.63 \times 10^{14}\text{ Hz}$
Solution: $\nu_0 = \dfrac{\phi_0}{h} = \dfrac{3.2 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 4.83 \times 10^{14}\text{ Hz}$.
Q15 — Work Function & Threshold · hard · numerical
Caesium has a work function of about $1.9\text{ eV}$. Its threshold wavelength is approximately:
A. $496\text{ nm}$
B. $1240\text{ nm}$
C. $653\text{ nm}$ ✓ Correct
D. $310\text{ nm}$
Solution: $\lambda_0 = \dfrac{1240}{1.9} \approx 653\text{ nm}$, which lies in the red part of the visible spectrum.
Q16 — Work Function & Threshold · hard · numerical
A metal has threshold wavelength $620\text{ nm}$. Its threshold frequency is approximately:
A. $2.07 \times 10^{14}\text{ Hz}$
B. $4.84 \times 10^{14}\text{ Hz}$ ✓ Correct
C. $4.84 \times 10^{15}\text{ Hz}$
D. $6.2 \times 10^{14}\text{ Hz}$
Solution: $\nu_0 = \dfrac{c}{\lambda_0} = \dfrac{3 \times 10^8}{620 \times 10^{-9}} \approx 4.84 \times 10^{14}\text{ Hz}$.
Q17 — Work Function & Threshold · easy · numerical
A work function of $2\text{ eV}$ expressed in joule is:
A. $3.2 \times 10^{-19}\text{ J}$ ✓ Correct
B. $3.2 \times 10^{-18}\text{ J}$
C. $2 \times 10^{-19}\text{ J}$
D. $1.6 \times 10^{-19}\text{ J}$
Solution: $2 \times 1.6 \times 10^{-19} = 3.2 \times 10^{-19}\text{ J}$.
Q18 — Work Function & Threshold · easy · numerical
Two metals have work functions $2\text{ eV}$ and $4\text{ eV}$. Photoemission occurs more readily from:
A. Neither of them
B. Both equally
C. The metal of work function $2\text{ eV}$ ✓ Correct
D. The metal of work function $4\text{ eV}$
Solution: A lower work function means a longer threshold wavelength, so a wider range of light can eject electrons.
Q19 — Work Function & Threshold · hard · numerical
Light of wavelength $600\text{ nm}$ falls on a metal of work function $2.5\text{ eV}$. The result is:
A. No photoemission, since the photon energy is below the work function ✓ Correct
B. Photoemission with kinetic energy $2.07\text{ eV}$
C. Photoemission with kinetic energy $0.4\text{ eV}$
D. Photoemission with kinetic energy $2.5\text{ eV}$
Solution: Photon energy $= \dfrac{1240}{600} \approx 2.07\text{ eV}$, which is less than $2.5\text{ eV}$.
Q20 — Work Function & Threshold · hard · numerical
A metal has threshold frequency $6 \times 10^{14}\text{ Hz}$. Its work function is approximately:
A. $9.05 \times 10^{-19}\text{ J}$
B. $6.63 \times 10^{-19}\text{ J}$
C. $3.98 \times 10^{-19}\text{ J}$ ✓ Correct
D. $1.99 \times 10^{-19}\text{ J}$
Solution: $\phi_0 = h\nu_0 = 6.63 \times 10^{-34} \times 6 \times 10^{14} \approx 3.98 \times 10^{-19}\text{ J}$, about $2.5\text{ eV}$.
Q21 — Work Function & Threshold · medium · numerical
A metal has work function $5\text{ eV}$. Its threshold wavelength is:
A. $496\text{ nm}$
B. $124\text{ nm}$
C. $248\text{ nm}$ ✓ Correct
D. $620\text{ nm}$
Solution: $\lambda_0 = \dfrac{1240}{5} = 248\text{ nm}$, in the ultraviolet.