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MCQEasyJEE 2023Photoelectric Effect

JEE Physics 2023 Question with Solution

The variation of stopping potential (V0V_0) as a function of the frequency (vv) of the incident light for a metal is shown in the figure. The work function of the surface is:

Graph of stopping potential V0 in volt versus frequency v in units of 10^14 Hz, showing a straight line starting near 5 × 10^14 Hz at zero stopping potential and rising to about 2.5 V near 10 × 10^14 Hz.
  • A

    2.07 eV2.07 \, \text{eV}

  • B

    18.6 eV18.6 \, \text{eV}

  • C

    2.98 eV2.98 \, \text{eV}

  • D

    1.36 eV1.36 \, \text{eV}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The graph shows stopping potential V0V_0 versus frequency vv for the incident light.

Find: The work function of the metal surface.

For photoelectric emission,

eV0=h(v−vth)eV_0 = h(v - v_{\text{th}})

At threshold frequency, stopping potential becomes zero, so

ϕ=hvth\phi = h v_{\text{th}}

From the graph, the threshold frequency is approximately

vth=5×1014 Hzv_{\text{th}} = 5 \times 10^{14} \, \text{Hz}

Therefore,

ϕ=(6.6×10−34)(5×1014)\phi = (6.6 \times 10^{-34}) (5 \times 10^{14}) ϕ=3.3×10−19 J\phi = 3.3 \times 10^{-19} \, \text{J}

Converting into electron volt,

ϕ=3.3×10−191.6×10−19 eV=2.07 eV\phi = \frac{3.3 \times 10^{-19}}{1.6 \times 10^{-19}} \, \text{eV} = 2.07 \, \text{eV}

Therefore, the work function is 2.07 eV2.07 \, \text{eV}. Although the solution labels the correct option as C, the worked calculation clearly gives 2.07 eV2.07 \, \text{eV}, which corresponds to option A.

Common mistakes

  • Using the slope of the graph to calculate the work function directly is wrong because the work function depends on the threshold frequency, not on the slope. First read the vv-axis intercept where V0=0V_0 = 0.

  • Reading the threshold frequency incorrectly from the graph leads to the wrong answer. The line starts near 5×1014 Hz5 \times 10^{14} \, \text{Hz}, so that value must be used for vthv_{\text{th}}.

  • Forgetting to convert joules to electron volts is incorrect because the options are given in eV. After finding ϕ\phi in joules, divide by 1.6×10−191.6 \times 10^{-19} to express it in electron volts.

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