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) = ${\displaystyle \frac{1.98\times {10}^{-19}}{1.6\times {10}^{-19}J/eV}=1.24eV}$

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 

${\displaystyle K.E=\frac{1}{2}m{V}_{max}^2}$

${\displaystyle V_{max}^2=\frac{2\times K.E}{m}}$

= ${\displaystyle \frac{2\times 3.98\times {10}^{-19}}{9.1\times {10}^{-31}}}$

= ${\displaystyle \frac{7.96}{9.1}\times {10}^{12}}$

V² = 87.4 × 10¹⁰ 

${\displaystyle V=\sqrt{87.4\times {10}^8}}$

= 9.3 × 10⁵ m/s