Verified Solution • 2025–26 Session moderate

QJEEM22Pha887: JEE Main 2022 Physics - Class 12 JEEM22Pha887 Solution: The stopping potential for photoelectrons emitted from a…

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 Ao\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)

Step-by-Step Verified Solution

hcλ−ϕ=KE=eV0{{hc} \over \lambda } - \phi = KE = e{V_0}

⇒6.63×10−34×3×1086630×10−10−6.63×10−34fth=1.6×10−19×0.4\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

⇒fth≃35.11×1013 H\Rightarrow {f_{th}} \simeq 35.11 \times {10^{13}}\,H

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