A metal (threshold frequency 8.0×1015Hz) is irradiated with the radiation of frequency 10.40×1015Hz. The stopping potential of the emitted electrons is about

A metal (threshold frequency $8.0 \times 10^{15} Hz$ ) is irradiated with the radiation of frequency $10.40 \times 10^{15} Hz$ . The stopping potential of the emitted electrons is about

A
67.6 V
B
82.1 V
C
120.1 V
D
99.4 V

Solution:

$\begin{aligned} | e V | = h v - h v_{0} \\ V = \frac{h (v - v_{0})}{e} = \frac{(6.626 \times 10^{- 34} Js) (2.40 \times 10^{16} s^{- 1})}{(1.6 \times 10^{- 19} C)} = 99.4 V \end{aligned}$

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Solution:

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