Question

Electrons ejected from the surface of a metal, when light of certain frequency is incident on it, are stopped fully by a retarding potential of 3 volts. Photo electric effect in this metallic surface begins at a frequency $6\times {10}^{14}{s}^{-1}$. The frequency of the incident light in $s^{-1}$ is :

($h=6\times {10}^{-34}$J-sec ; charge on the electron $=$ $1.6\times {10}^{-19}C$ )

• A. $7.5\times {10}^{13}$
• B. $13.5\times {10}^{13}$
• C. $14\times {10}^{14}$
• D. $7.5\times {10}^{15}$

Answer 1

The correct option is C $14\times {10}^{14}$

Stopping potential $=(h\nu -h{\nu }_0)/e$

$3\times 1.6\times {10}^{-19}=6\times {10}^{-34}(\nu -{\nu }_0)(e=1.6\times {10}^{-19}c)$

$\frac{3\times 1.6\times {10}^{-19}}{6\times {10}^{-34}}=\nu -6\times {10}^{14}$

$\nu =6\times {10}^{14}+\frac{3\times 1.6\times {10}^{-19}}{6\times {10}^{-34}}$

$=6\times {10}^{14}+0.8\times {10}^{15}$

$=14\times {10}^{14}se{c}^{-1}$

So, the answer is (C).