Question

Threshold frequency of a metal is $5\times {10}^{13}se{c}^{-1}$ upon which $1\times {10}^{14}se{c}^{-1}$ frequency light focused then maximum kinetic energy of emitted electron?

• A. $3.3\times {10}^{-21}$
• B. $3.3\times {10}^{-20}$
• C. $6.6\times {10}^{-21}$
• D. $6.6\times {10}^{-20}$

Answer 1

The correct option is B $3.3\times {10}^{-20}$

Frequency of focused light = $1\times {10}^{14}=10\times {10}^{13}$

$\varphi =$ threshold frequency = $5\times {10}^{13}$

$K.E=hv-h\varphi$

$=h(10-5)\times {10}^{13}=(6.626\times {10}^{-34})\times 5\times {10}^{13}$

$=3.3\times {10}^{-20}$

So, the correct option is $B$