Question

A metal surface of work function 2.2ev is when eradicated with light as ejected out of electron of velocity 6×10^6m/s what would have been the wavelength in nm of incident photon

Answer 1

Work function , Ф = 2.2 eV = 2.2 × 1.6 × 10⁻¹⁹ J = 3.52 × 10⁻¹⁹ J
Velocity of ejected electron , v = 6 × 10⁶ m/s
mass of electron = 9.1 × 10⁻³¹ Kg
Let wavelength of incident photon is λ

Now, use photoelectric effect formula ,
kinetic energy = energy of incident photon - energy of ejected electron
1/2mv² = hc/λ - 3.52 × 10⁻¹⁹
1/2 × 9.1 × 10⁻³¹ × (6 × 10⁶)² = 6.6 × 10⁻³⁴ × 3 × 10⁸/λ - 3.52 × 10⁻¹⁹
9.1 × 18 × 10⁻¹⁹ + 3.52 × 10⁻¹⁹ = 19.8 × 10⁻²⁶/λ
(163.8 + 3.52) × 10⁻¹⁹ = 19.8 × 10⁻²⁶/λ
167.32 × 10⁻¹⁹ = 19.8 × 10⁻²⁶/λ
λ = 198 × 10⁻²⁷/167.32 × 10⁻¹⁹
=1.183 × 10⁻⁸ m