Question

The threshold frequency fora certain metal is 3.3 × 10¹⁴ Hz. If light of frequency 8.2 × 10¹⁴ Hz is incident on the metal, predict the cuto voltage for the photoelectric emission.

Answer 1

Given v₀ = 3.3 × 10¹⁴ Hz

v = 8.2 × 10¹⁴ Hz

Since eV₀ = hv – hv₀

So V₀ = \(\frac{h}{e}\) (v-v₀)

= \(\frac{6.62\times10^{-34}}{1.6\times10^{-19}}\) x (8.2 x 10¹⁴ - 3.3 x 10¹⁴)

= 4.14 × 10^-15 × 4.9 × 10¹⁴

= 2.02 V = 2.0 V