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MCQEasyJEE 2024Photoelectric Effect

JEE Physics 2024 Question with Solution

The threshold frequency of a metal with work function 6.63 eV6.63 \, \text{eV} is:

  • A

    16×1015 Hz16 \times 10^{15} \, \text{Hz}

  • B

    16×1012 Hz16 \times 10^{12} \, \text{Hz}

  • C

    1.6×1012 Hz1.6 \times 10^{12} \, \text{Hz}

  • D

    1.6×1015 Hz1.6 \times 10^{15} \, \text{Hz}

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: Work function ϕ=6.63 eV\phi = 6.63 \, \text{eV}.

Find: Threshold frequency f0f_0.

Use Einstein's photoelectric relation:

ϕ=hf0\phi = h f_0

where h=4.1357×10−15 eV⋅sh = 4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}.

Substituting the given values:

f0=ϕh=6.634.1357×10−15 Hzf_0 = \frac{\phi}{h} = \frac{6.63}{4.1357 \times 10^{-15}} \, \text{Hz}

On calculation:

f0=1.602×1015 Hzf_0 = 1.602 \times 10^{15} \, \text{Hz}

Rounding gives 1.6×1015 Hz1.6 \times 10^{15} \, \text{Hz}.

Therefore, the threshold frequency is 1.6×1015 Hz1.6 \times 10^{15} \, \text{Hz}. The solution states the correct option is B, but this value matches option D in the listed options. Hence there is an answer-key discrepancy, and the defensible option from the given options is D.

Using SI Units

Given: ϕ0=6.63 eV\phi_0 = 6.63 \, \text{eV} and h=6.63×10−34 J⋅sh = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s}.

Find: ν0\nu_0.

First convert electron volt into joule:

ϕ0=6.63×1.6×10−19 J=1.06×10−18 J\phi_0 = 6.63 \times 1.6 \times 10^{-19} \, \text{J} = 1.06 \times 10^{-18} \, \text{J}

Now apply:

ϕ0=hν0\phi_0 = h \nu_0

So,

ν0=ϕ0h=1.06×10−186.63×10−34\nu_0 = \frac{\phi_0}{h} = \frac{1.06 \times 10^{-18}}{6.63 \times 10^{-34}}

Thus,

ν0≈1.6×1015 Hz\nu_0 \approx 1.6 \times 10^{15} \, \text{Hz}

Therefore, the threshold frequency is 1.6×1015 Hz1.6 \times 10^{15} \, \text{Hz}. This again matches option D by value, although the solution labels it as B.

Common mistakes

  • Using the option label from the solution without checking the numerical value is incorrect because the solution contains a mismatch between label and value. Match the computed frequency with the listed options before choosing the answer.

  • Using h=6.63×10−34h = 6.63 \times 10^{-34} without converting the work function from electron volt to joule is incorrect because the units become inconsistent. Either use hh in eV⋅s\text{eV} \cdot \text{s} or convert ϕ\phi into joule first.

  • Writing the threshold relation incorrectly as involving emitted electron energy is wrong because at threshold the kinetic energy is zero. Use ϕ=hf0\phi = h f_0 directly.

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