Question

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

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

Answer 1

Answer: (D)

$d=0.24mm,D=1.2m$
fringe width

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

$0.002=\frac{1.2\lambda }{.00024}$

Thus we get
$\lambda =4\times {10}^{-7}m$

Thus the energy of the light is given as $\frac{hc}{\lambda }=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{4\times {10}^{-7}}=4.97\times {10}^{-19}Joules$
or

$\frac{4.97\times {10}^{-19}}{1.6\times {10}^{-19}}=3.1eV$

The work function is given as 2.2 eV
the stopping potential thus required is $3.1-2.2=0.9eV$

So, the answer is option (C).