Question

A light of wavelength 400 nm is incident on metal surface whose work function is 3.0 × 10⁻¹⁹ J. Calculate the speed of the fastest photoelectrons emitted.

Answer 1

Given: h = 6.63 × 10⁻³⁴ J.s

c = 3.0 × 10⁸ m/s

λ = 400 nm = 400 × 10⁻⁹ m

Φ = 3.0 × 10⁻¹⁹ J

Using the photoelectric equation:

KE_max = hf − Φ

= hc/λ − Φ

= `(6.63 xx 10^-34) ((3.0 xx 10^8))/(400 xx 10^-9) - 3.0 xx 10^19 J`

= 4.97 × 10⁻¹⁹ − 3.0 × 10⁻¹⁹

= 1.97 × 10¹⁹ J

Using KE = `1/2 mv^2`

Solving for v:

v = `sqrt(2KE//m_c)`

= `sqrt(2((1.97 xx 10^-19)/(9.11 xx 10^-1)))`

≈ 6.6 × 10⁵ m/s