the de broglie wavelength of the electron ejected out when the radiation of energy 'E' strikes on the metal surface having threshold frequency equal to half of incident frequency is?

Dear Student,

Let the Work function be W, incident energy be E and the kinetic energy of the emitted electron be K.
So,

$$\begin{gathered} E = h\nu , h = planck\text{'}s cons\tan t ,\nu = frequency \\ W = \frac{h\nu }{2} \\ K = \frac{1}{2}m{u}^2, m = mass of electron, u = velocity \\ So, \\ By photoelectric effect, \\ E = W + K \\ K = E - W = h\nu -\frac{h\nu }{2} = hv = \frac{E}{2} \\ Now, \\ K =\frac{E}{2}= \frac{1}{2}m{u}^2 \\ 2 \times \frac{E}{2} = m{u}^2 \\ Em =m^2{u}^2=p^2 , where p = momentum \\ So, p = \sqrt{Em} \end{gathered}$$.



Now, According to De Broglie's Equation,

$$\begin{gathered} \lambda = \frac{h}{p}, where \lambda =De Broglie\text{'}s wavelength of the emitted electron \\ So, from above, \\ \lambda =\frac{h}{\sqrt{Em}} \end{gathered}$$
 

Hope it helps.

Regards