JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Modern Physics

When radiation of wavelength

λ

is incident on a metallic surface, the stopping potential of ejected photoelectrons is 4.8 V. If the same surface is illuminated by radiation of double the previous wavelength, then the stopping potential becomes 1.6 V. The threshold wavelength of the metal is:

(A) $6 λ$
(B) $8 λ$
(C) $2 λ$
(D) $4 λ$ ✓ Correct

Answer: (D)

Solution

The photoelectric equation gives

e V_{s} = \frac{h c}{λ} - ϕ

. For the two wavelengths,

4.8 e = \frac{h c}{λ} - ϕ, 1.6 e = \frac{h c}{2 λ} - ϕ

Subtracting the second from the first,

3.2 e = \frac{h c}{2 λ}

, so

\frac{h c}{λ} = 6.4 e

and

ϕ = 6.4 e - 4.8 e = 1.6 e

. The threshold wavelength satisfies

ϕ = \frac{h c}{λ _{0}}

, hence

λ_{0} = \frac{h c}{1.6 e} = \frac{6.4 e λ}{1.6 e} = 4 λ

The threshold wavelength of the metal is

4 λ

JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Single correct option

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JEE Main · 2021 · 25 Jul 2021 · Shift 2 · Modern Physics

Solution

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