Question

For photoelectric effect in a metal, the graph of stopping potential $V_0\left(in\text{V}\right)$ versus frequency $\nu \left(in\text{Hz}\right)$ of the incident radiation as shown in figure. From the graph, find threshold frequency and work function of the metal. (Take $h=6.6\times {10}^{-34}\text{J s}$)

• A. $4\times {10}^{15}\text{Hz};4\text{eV}$
• B. $4\times {10}^{15}\text{Hz};16.5\text{eV}$
• C. $16.5\times {10}^{15}\text{Hz};4\text{eV}$
• D. $8\times {10}^{15}\text{Hz};12\text{eV}$

Answer 1

The correct option is B $4\times {10}^{15}\text{Hz};16.5\text{eV}$

From, the Einstein's photoelectric equation,

$E={\varphi }_0+K{E}_{max}$

$\Rightarrow E-{\varphi }_0=e{V}_0$

Where, ${\varphi }_0=\text{Work function}=h{\nu }_0$

$\left(\text{a}\right)$ From, the figure we get, the threshold frequency, ${\nu }_0=4\times {10}^{15}\text{Hz}$

$\left(\text{b}\right){\varphi }_0=h{\nu }_0=(6.6\times {10}^{-34})\times (4\times {10}^{15})$

$=26.4\times {10}^{-19}\text{J}\text{(or)}$

$=\frac{26.4\times {10}^{-19}}{1.6\times {10}^{-19}}=16.5\text{eV}$

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

Why this question? Key Notes: In the graph, threshold frequency is taken at a point where, stopping potential is zero.