Question

In the arrangement shown in figure, $y=1.0\text{mm},d=0.24\text{mm}$ and $D=1.2\text{m}$. The work function of the material of the emitter is $2.2\text{eV}.$ Find the stopping potential $V$ needed to stop the photocurrent.
[Use $hc=1240\text{eV (nm)}$]

• A. $0.9\text{V}$
• B. $3.1\text{V}$
• C. $5.3\text{V}$
• D. Zero

Answer 1

The correct option is A $0.9\text{V}$

Given,

$y=1.0\text{mm};d=0.24\text{mm}D=1.2\text{m};{\varphi }_0=2.2\text{eV}$

From the YDSE, position of the first order bright fringe is,

$y=\frac{\lambda D}{d}$

$\lambda =\frac{yd}{D}=\frac{2\times {10}^{-3}\times 0.24\times {10}^{-3}}{1.2}=4\times {10}^{-7}\text{m}\text{(or)}$

$\lambda =400\times {10}^{-9}\text{m}=400\text{nm}$

The energy of the incident radiation is,

$E=\frac{hc}{\lambda }=\frac{1240\text{eV (nm)}}{400}=3.1\text{eV}$

From, Einstein's photoelectric equation,

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

$\Rightarrow V_0=3.1-2.2=0.9\text{V}$

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