De Broglie wavelength of photoelectrons is $1\: A^{\circ}$ . What is stopping potential?

Question

$\begin {array} {1 1} (a)\;151\: V & \quad (b)\;153\: V \\ (c)\;123\: V & \quad (d)\;136\: V \end {array}$

 

Answer 1

Ans : (a)

$E_{max} = eV_{\circ}$

So, $ \lambda_e = \large\frac{h}{\sqrt{2mE_{max}}} = \large\frac{h}{\sqrt{2meV_{\circ}}}$

$ \Rightarrow V_{\circ} =\large\frac{ h^2} { 2me \lambda_e^2}$

$ \Rightarrow V_{\circ} =\large\frac{ (6.6 \times 10^{-34} )^2}{ [2 \times (9.1 \times 10^{-31} ) \times 1.6 \times 10^{-19} \times (10^{-10} ) ^2}$

$ \Rightarrow V_{\circ} = 151\: V$