When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ____ m. (h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is

Q. When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is ____ m. (h = 6.63 × 10⁻³⁴ J·s, e = 1.6 × 10⁻¹⁹ C)

(A) 2.9 × 10⁻⁸

(B) 2.5 × 10⁻⁷

(D) 2.2 × 10⁻⁸

Correct Answer: 2.5 × 10⁻⁷ m

Explanation

According to Einstein’s photoelectric equation,

hν = φ + eV_s

For the first light of wavelength λ:

h c / λ = φ + e(3.2)

For the second light, wavelength is doubled i.e. 2λ and stopping potential is 0.7 V:

h c / (2λ) = φ + e(0.7)

Subtracting the second equation from the first:

h c / λ − h c / (2λ) = e(3.2 − 0.7)

h c / (2λ) = 2.5 e

λ = h c / (5e)

Substituting values:

λ = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (5 × 1.6 × 10⁻¹⁹)

λ = 1.989 × 10⁻²⁵ / 8 × 10⁻¹⁹

λ ≈ 2.5 × 10⁻⁷ m

Hence, the wavelength of the first light is 2.5 × 10⁻⁷ m.

Explanation

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