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$-$34 Js)
Answer (integer)
35
Solution
<p>${{hc} \over \lambda } - \phi = KE = e{V_0}$</p>
<p>$$ \Rightarrow {{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}} \over {6630 \times {{10}^{ - 10}}}} - 6.63 \times {10^{ - 34}}{f_{th}} = 1.6 \times {10^{ - 19}} \times 0.4$$</p>
<p>$\Rightarrow {f_{th}} \simeq 35.11 \times {10^{13}}\,H$</p>
About this question
Subject: Physics · Chapter: Dual Nature of Matter and Radiation · Topic: Photoelectric Effect
This question is part of PrepWiser's free JEE Main question bank. 145 more solved questions on Dual Nature of Matter and Radiation are available — start with the harder ones if your accuracy is >70%.