A photo-emissive substance is illuminated with a radiation of wavelength λ1 so that it releases electrons

Question

A photo-emissive substance is illuminated with a radiation of wavelength \(\lambda _i\) so that it releases electrons with de-Broglie wavelength \(\lambda _e\). The longest wavelength of radiation that can emit photoelectron is \(\lambda _0\). Expression for de-Broglie wavelength is given by \(:\)

(m \(:\) mass of the electron, h \(:\) Planck’s constant and c \(:\) speed of light)

(1) \(\lambda _e = \sqrt{\frac{h}{2mc \left( \frac{1}{\lambda_i} -\frac{1}{\lambda_0}\right)}}\)

(2) \(\lambda _e = \sqrt{\frac{h\lambda _0}{2mc}}\)

(3) \(\lambda _e =\frac{h}{\sqrt{2mc \left( \frac{1}{\lambda _i} - \frac{1}{\lambda _0}\right)}}\)

(4) \(\lambda _e = \sqrt{\frac{h\lambda_i}{2mc}}\)

Answer 1

Correct option is : (1) \(\lambda _e = \sqrt{\frac{h}{2mc \left( \frac{1}{\lambda_i} -\frac{1}{\lambda_0}\right)}}\)

K.E = E – W

\(\lambda _e = \frac{h}{\sqrt{2m K.E}}, E = \frac{hc}{\lambda _1}, W = \frac{hc}{\lambda _0}\)

\(\frac{h^2}{2m\lambda _e^2} = \frac{hc}{\lambda _i} - \frac{hc}{\lambda _0}\)

\(\lambda _e = \sqrt{\frac{h}{2mc \left(\frac{1}{\lambda _i} - \frac{1}{\lambda _0}\right)}}\)