Question:

Surface of certain metal is first illuminated with light of wavelength λ₁=350nm and then, by light of wavelength λ₂=54nm. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to: (Energy of photon=1240/λ(in nm)eV)

5.6

1.4

1.8

2.5

We know that,

(hc/λ₁) = Φ + (1/2)m(2v)²
(hc/λ₂) = Φ + (1/2)mv²

⇒ (hc/λ₁) - Φ = 4[(hc/λ₂) - Φ]

Solving this equation for Φ, we get:

Φ = (4hc/λ₂) - (hc/λ₁)
= hc(4/λ₂ - 1/λ₁)
= 1240(4/54 - 1/350) eV
≈ 1.8 eV