Question

Threshold frequency, $v_0$ is the minimum frequency which a photon must possess to eject an electron from a metal. It is different for different metals. When a photon of frequency $1.0\times {10}^{15}{S}^{-1}$ was allowed to hit a metal surface, an electron having $1.988\times {10}^{-19}J$ of kinetic energy was emitted. The threshold frequency of this metal is:

• A. $7\times {10}^{14}{s}^{-1}$
• B. $6\times {10}^{14}{s}^{-1}$
• C. $8\times {10}^{14}{s}^{-1}$
• D. $9\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is A $7\times {10}^{14}{s}^{-1}$

$h\vartheta =h{\vartheta }_0+K.E$

$K.E=1.988\times {10}^{-19}$ J

$\vartheta =1\times {10}^{15}$ $s^{-1}$

h ( plank constant) $=6.626\times {10}^{-34}$ Js

${\vartheta }_0=(h\nu -K.E)/h$

${\vartheta }_0=(6.626\times {10}^{-34}\times 1\times {10}^{15}-1.988\times {10}^{-19})/6.626\times {10}^{-34}$

$=7\times {10}^{14}$ $s^{-1}$