A light of wavelength ( < ₀) falls on a metal surface. The momentum of the fastest moving electron is p . In order to have the momentum of an electron equal to 2p , the required wavelength ' of incident light must be

Step-by-step solution

Using Einstein's photoelectric equation: K_ = hc - hc ₀ Since K_ = p^2 2m , we have: p^2 2m = hc - hc ₀ For the new wavelength ' , the momentum is 2p . The new maximum kinetic energy is: K'_ = (2p)^2 2m = 4 ( p^2 2m ) Applying the photoelectric equation again: 4 ( p^2 2m ) = hc ' - hc ₀ Substituting the value of p^2 2m from the first equation: 4 ( hc - hc ₀ ) = hc ' - hc ₀ Dividing both sides by hc : 4 - 4 ₀ = 1 ' - 1 ₀ 1 ' = 4 - 3 ₀ 1 ' = 4 ₀ - 3 ₀ ' = ₀ 4 ₀ - 3 Answer: ₀ 4 ₀ - 3