Question

In an experiment of photoelectric effect, a metal produces electron with frequency of 5.2 × 10¹⁴ s^-1 and maximum kinetic energy of 1.3 x 10^-19 J. What will be threshold frequency of the metal? 

Answer 1

As we know, hv =  hv° + K.E 

hv - hv° = K.E 

h (v - v°) = K.E or (v - v°) \( = \frac{K.E}{h}\)

\( = 5.2\times 10^{14}\frac{1.3\times 10^{-19}}{6.626\times 10^{-34}}\)

= 5.2 × 10¹⁴ -1.96 × 10¹⁴ 

= 3.24 × 10¹⁴s^-1