The threshold frequency of a metal with work function 6.63 eV is: — Solution | JEE Physics

MCQEasyJEE 2024Photoelectric Effect

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

A
$16 \times 10^{15} \, \text{Hz}$
B
$16 \times 10^{12} \, \text{Hz}$
C
$1.6 \times 10^{12} \, \text{Hz}$
D
$1.6 \times 10^{15} \, \text{Hz}$

Answer

Correct answer:D

Step-by-step solution

Standard Method

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

Find: Threshold frequency $f_0$ .

Use Einstein's photoelectric relation:

\phi = h f_0

where $h = 4.1357 \times 10^{-15} \, \text{eV} \cdot \text{s}$ .

Substituting the given values:

f_0 = \frac{\phi}{h} = \frac{6.63}{4.1357 \times 10^{-15}} \, \text{Hz}

On calculation:

f_0 = 1.602 \times 10^{15} \, \text{Hz}

Rounding gives $1.6 \times 10^{15} \, \text{Hz}$ .

Therefore, the threshold frequency is $1.6 \times 10^{15} \, \text{Hz}$ . 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: $\phi_0 = 6.63 \, \text{eV}$ and $h = 6.63 \times 10^{-34} \, \text{J} \cdot \text{s}$ .

Find: $\nu_0$ .

First convert electron volt into joule:

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

Now apply:

\phi_0 = h \nu_0

So,

\nu_0 = \frac{\phi_0}{h} = \frac{1.06 \times 10^{-18}}{6.63 \times 10^{-34}}

Thus,

\nu_0 \approx 1.6 \times 10^{15} \, \text{Hz}

Therefore, the threshold frequency is $1.6 \times 10^{15} \, \text{Hz}$ . 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 $h = 6.63 \times 10^{-34}$ without converting the work function from electron volt to joule is incorrect because the units become inconsistent. Either use $h$ in $\text{eV} \cdot \text{s}$ or convert $\phi$ into joule first.
Writing the threshold relation incorrectly as involving emitted electron energy is wrong because at threshold the kinetic energy is zero. Use $\phi = h f_0$ directly.

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Answer

Step-by-step solution

Common mistakes

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