Question

If work function of an arbitrary metal is 3.1 EV find its threshold wavelength when radiation of 300 nm strike the metal surface?​

Answer 1

Given:

Work Function of metal, W = 3.1 eV = 3.1 × 1.6 × 10⁻¹⁹ J

Wavelength of light, λ = 300 nm = 3 × 10⁻⁷ m

To Find:

The threshold wavelength (λ₀) of the metal.

Calculation:

- We know that the energy is given as:

E = E₀ + W

⇒ hc / λ = (hc / λ₀) + W

⇒ (6.63 × 10⁻³⁴ × 3 × 10⁸)/ (3 × 10⁻⁷) = (6.63 × 10⁻³⁴ × 3 × 10⁸)/λ₀ + (3.1 × 1.6 × 10⁻¹⁹)

⇒ 6.63 × 10⁻¹⁹ = (19.89 × 10⁻²⁶/ λ₀) + (4.96 × 10⁻¹⁹)

⇒ 19.89 × 10⁻²⁶/ λ₀ = (6.63 - 4.96) × 10⁻¹⁹

⇒ 19.89 × 10⁻²⁶/ λ₀ = 1.67 × 10⁻¹⁹

⇒ λ₀ = 19.89 × 10⁻²⁶ / 1.67 × 10⁻¹⁹

⇒ λ₀ = 11.91 × 10⁻⁷

⇒ λ₀ = 1191 × 10⁻⁹ m = 1191 nm

- So, the threshold wavelength is 1191 nm.