In a photoelectric experiment, ultraviolet light of wavelength $280nm$ is used with lithium cathode having work function $ϕ = 2 . 5eV$ . If the wavelength of incident light is switched to $400nm$ , find out the change in the stopping potential.
\left(h = 6 . 63 \times {10}^{- 34}Js, c = 3 \times {10}^{8}{ms}^{- 1}\right)$

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In a photoelectric experiment, ultraviolet light of wavelength $280nm$ is used with lithium cathode having work function $ϕ = 2 . 5eV$ . If the wavelength of incident light is switched to $400nm$ , find out the change in the stopping potential.

\left(h = 6 . 63 \times {10}^{- 34}Js, c = 3 \times {10}^{8}{ms}^{- 1}\right)$

Quantrex Academy · Accepted Answer

Correct answer: C. 1 . 3V

From the photoelectric equation, e V _ 0 = (hc)/(λ) - ϕ Using this equation for the given two cases, e V _ 1 = (hc)/(280nm) - ϕ e V _ 2 = (hc)/(400nm) - ϕ On subtracting, e( V _ 1 - V _ 2 ) = 1240((120)/(280 × 400)) = 1.3eV ⇒ ( V _ 1 - V _ 2 ) = 1 . 3V

In a photoelectric experiment, ultraviolet light of wavelength...

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