Question

In a photoelectric emission, electrons are ejected from metals $\text{P}$ and $\text{Q}$ by light of frequency $f$. The potential difference $V$ required to stop the electrons is measured for various frequencies. If $\text{Q}$ has a greater work function, then $\text{P}$ which graph illustrates the expected results?

• A.
• B.
• C.
• D.

Answer 1

The correct option is D

Electrons gets ejected, $$\text{if},\{\begin{array}{cc}h& {\nu }_{th}>\varphi \\ {\nu }_{th}& >\frac{\varphi }{h}\end{array}$$

Where, ${\nu }_{th}=\text{Threshold frequency}\varphi =\text{Work function}$

The graph between $Vvsf$ has a constant slope of $\left(\frac{h}{e}\right)$, so both lines must be parallel.

But the work function $\text{Q}$ is greater than $\text{P}$ [i.e., Threshold frequency of $\text{Q}$ is greater than $\text{P}$]

$\therefore {\varphi }_Q>{\varphi }_P\Rightarrow ({\nu }_{th}{)}_Q>({\nu }_{th}{)}_P$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.