Question

In an experiment on photoelectric emission from a metallic surface, wavelength of incident light is $2\times {10}^{-7}m$ and stopping potential is 2.5V. The threshold frequency of the metal in Hz. Approximately (e) $=1.6\times {10}^{-19}C$,$h=6.6\times {10}^{-34}$Js.

• A. $12\times {10}^{15}$
• B. $9\times {10}^{15}$
• C. $9\times {10}^{14}$
• D. $12\times {10}^{13}$

Answer 1

Answer: (C)

$9\times {10}^{14}$

$\lambda =2\times {10}^{-7}m=2000A^o=200nm$
stopping potential = $(\frac{hc}{\lambda }-\varphi )/e$
$2.5V\times e=\frac{hc}{\lambda }-\varphi$

$2.5eV=\frac{1240}{200}-\varphi$ $(hc=1240eV-nm)$

$\varphi =(6.2-2.5)eV$
$=3.7eV$
$\varphi =h{V}_0$
$3.7\times 1.6\times {10}^{-19}=6.6\times {10}^{-34}{V}_0$

$V_0=\frac{3.7\times 1.6\times {10}^{-19}}{6.6\times {10}^{-34}}$

$=0.9\times {10}^{15}$
$=9\times {10}^{14}Hz$

So, the answer is option (C).