JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 26)
The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $$\mathop A\limits^o $$ is 0.42 V. If the threshold frequency is x $$\times$$ 1013 /s, where x is _________ (nearest integer).
(Given, speed light = 3 $$\times$$ 108 m/s, Planck's constant = 6.63 $$\times$$ 10$$-$$34 Js)
Answer
35
Explanation
$${{hc} \over \lambda } - \phi = KE = e{V_0}$$
$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6630 \times {{10}^{ - 10}}}} - 6.63 \times {10^{ - 34}}{f_{th}} = 1.6 \times {10^{ - 19}} \times 0.4$$
$$ \Rightarrow {f_{th}} \simeq 35.11 \times {10^{13}}\,H$$
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