Question

If photoelectrons are to be emitted from a potassium surface with a speed 6 x 10⁶ ms^-1, what frequency of radiation must be used? (Threshold frequency for potassium is 4.22 x 10¹⁴ Hz, h = 6.6 x 10^-34Js, me = 9.1 x 10^-31kg)

Answer 1

Here, v = 6 x 10⁶ ms^-1 

V₀ = 4.22 x 10¹⁴ Hz 

From Einstein’s photoelectric equation, 

k.E = \(\frac{1}{2}\)mv² = h (υ – υ0)

= (2.48 x 10¹⁴) + (4. 22 x 10^-14 ) 

υ = 6.7 x10¹⁴ Hz