The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 A o is 0.42 V. If the threshold frequency is x * 1013 /s, where x is ________(nearest integer). (Given, speed light = 3 * 108 m/s, Planck's constant = 6.63 * 10-34 Js)

eV0+hυ0=hCλ ⇒1.6*10−19*0.42+6.63*10−34υ0=6.63*10−34*3*1086630*10−10 ⇒6.63*10−34υ0=10−19 (3−1.6*0.42) 6.63*10−34υ0=2.328*10−19 υ0=35.11*1013=35

The stopping potential for photoelectrons emitted from a surface illuminated by light of wavelength 6630 $\overset{o}{A}$ **is 0.42 V. If the threshold frequency is x * 10¹³ /s, where x is ________(nearest integer).

(Given, speed light = 3 * 10⁸ m/s, Planck's constant = 6.63 * 10^-34 Js)**

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Vishal Baghel | Contributor-Level 10

8 months ago

$e V_{0} + h υ_{0} = \frac{h C}{λ}$

$\Rightarrow 1.6 * 1 0^{- 19} * 0.42 + 6.63 * 1 0^{- 34} υ_{0} = \frac{6.63 * 1 0^{- 34} * 3 * 1 0^{8}}{66 3^{0} * 1 0^{- 10}}$

$\Rightarrow 6.63 * 1 0^{- 34} υ_{0} = 1 0^{- 19} (3 - 1.6 * 0.42)$

$6.63 * 1 0^{- 34} υ_{0} = 2.328 * 1 0^{- 19}$

$υ_{0} = 35.11 * 1 0^{13} = 35$

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