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

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

 

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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$

 

answered Dec 27, 2013 by thanvigandhi_1
edited Mar 13, 2014 by thanvigandhi_1
 

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