Question

Light of frequency 7.21 × 10¹⁴ Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10⁵ m/s are ejected from the surface. What is the threshold frequency for photo emission of electrons?

Answer 1

Given v = 7.21 × 10¹⁴ Hz

u_max = 6.0 × 10¹⁴ ms^-1

v₀ = ?

Since K_max = hv – hv₀

∴ V₀ = v - K_max

_{= V - \(\frac{mv_{max}}{2h}\)}

= 7.21 x 10¹⁴ -\(\Big(\frac{9.1\times10^{-31}\times36\times10^{10}}{2\times6.62\times10^{-19}}ev\Big)\)

= 7.21 x 10¹⁴ - 2.48 x 10¹⁴ = 4.73 x 10¹⁴ Hz