Question

A metal surface is illuminated by light of two different wavelengths 207 nm and 414 nm. The maximum speeds of photoelectrons corresponding to these wavelengths are u₂ and u₂ respectively with u₁ : u₂ = 2 : 1. The work function of the metal is ______.

(hc =1242 eV nm)

Options

• 1.6 eV

• 2.0 eV

• 2.4 eV

• 3.0 eV

Answer 1

A metal surface is illuminated by light of two different wavelengths 207 nm and 414 nm. The maximum speeds of photoelectrons corresponding to these wavelengths are u₂ and u₂ respectively with u₁ : u₂ = 2 : 1. The work function of the metal is 2.0 eV.

Explanation:

`1/2 mv_1^2 = (hc)/lambda_1 - Phi_0`

`1/2 mv_2^2 = (hc)/lambda_2 - Phi_0`

`(v_1/v_2)^2 = (((hc)/lambda_1 - Phi_0)/((hc)/lambda_2 - Phi_0)) = (2/1)^2    ...[because v_1/v_2 = 2]`

∴ `(4 hc)/lambda_2 - 4Phi_0 = (hc)/lambda_1 - Phi_0`

∴ `(4 hc)/lambda_2 - (hc)/lambda_1` = 3Φ₀

∴ `(4 xx 1242)/414 - 1242/207` = 3Φ₀

∴ 3Φ₀ = 6

⇒ Φ₀ = 2 eV