Question

The threshold frequency for a metal is 3.0 × 10¹⁴ Hz. A beam of frequency 9.0 × 10¹⁴ Hz is incident on the metal. Calculate

1. the work function (in eV) of the metal and
2. the maximum speed of photoelectrons.

Answer 1

Threshold frequency (V₀) = 3.0 × 10¹⁴ Hz

Incident frequency (v) = 9.0 × 10¹⁴ Hz

i. Work function(Φ) = hv₀

h = 6.63 × 10³⁴ Js

Φ = 6.63 × 10⁻³⁴ × 3 × 10¹⁴ Hz

= 1.98 × 10⁻¹⁹ J, 1 eV = 1.6 × 10⁻¹⁹ J

Φ (in eV) = `(1.98 xx 10^-19)/(1.6 xx 10^-19 J//eV) = 1.24 eV`

ii. E = Φ + K.E.

hv = Φ + K.E

K.E = hv − Φ

= 6.63 × 10⁻³⁴ × 9 × 10¹⁴ − 1.98 × 10⁻¹⁹ 

= 5.96 × 10⁻¹⁹ J − 1.98 × 10⁻¹⁹ 

= 3.98 × 10⁻¹⁹ 

`K.E = 1/2 mV_(max)^2`

`V_(max)^2 = (2 xx K.E)/m`

= `(2 xx 3.98 xx 10^-19)/(9.1 xx 10^-31)`

= `7.96/9.1 xx 10^12`

V² = 87.4 × 10¹⁰ 

`V = sqrt(87.4 xx 10^8)`

= 9.3 × 10⁵ m/s