Question

A metal whose work function is 2 eV is illuminated by light of wavelength 3 × 10^-7 m. Calculate

1. threshold frequency

2. maximum energy of photo electrons

3. the stopping potential.

Answer 1

F₀ = h ν₀ – 2eV

v = \(\frac{c}{v}= \frac{3\times10^8}{3\times10^{-7}}\) = 10¹⁵Hz

1. F₀ = hv₀

v₀ = \(\frac{2\times1.6\times10^{-19}}{6.63\times10^{-34}}\)

= 4.8 x 10¹⁴ Hz

2.  1/2mv² = h(ν – ν₀)

= 6.63 × 10^-34 (10¹⁵ – 4.8 × 10¹⁴)

= 3.44 × 10^-19 J

3. eV₀ = h(ν – ν₀)

= V₀ = \(\frac{h(v-v_0)}{e}\)

= \(\frac{3.44\times10^{-19}}{1.6\times10^{-19}}\) = 5.746V