Question

The fringe width in a YDSE is 2 $mm$, distance between slits and screen 1.2 $m$ and separation between the slits is 0.24 $mm$. The radiation of same source is also incident on a photo cathode of work function 2.2 $eV$. Find the stopping potential.

• A. 3.1 $V$
• B. 2.2 $V$
• C. 0.9 $V$
• D. 5.3 $V$

Answer 1

Answer: (C)

0.9 $V$

We have fringe width,

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

$⟹\lambda =\frac{\beta d}{D}=\frac{2\times {10}^{-3}\times 0.24\times {10}^{-3}}{1.2}=400nm$

Also

$E=e{V}_s=(\frac{hc}{\lambda }-\varphi )J$

$⟹V_s=(\frac{1240}{\lambda }-\varphi )eV$

$⟹V_s=\frac{1240}{400\times {10}^{-9}}-2.2=3.1-2.2=0.9eV$

So, the answer is option (C).