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# De-Broglie proposed dual nature for electron by putting his famous equation $\lambda = \large\frac{h}{mu}$. Later on Heisenburg proposed uncertainity principle as $\bigtriangleup p .\bigtriangleup x \geq \large\frac{h}{2}(h = \large\frac{h}{2\pi})$ . On the contrary particle nature of electron was established on the basis of photoelectric effect. When a photon strikes the metal surface , it gives up its energy to the electron. Part of this energy (say w) is used by the electron to escape from the metal and the remaining imparts the kinetic energy $\large\frac{1}{2}mu^2$ to the photoelectron. The potential applied on the surface to reduce the velocity of photoelectron to zero is known as stopping potential. With what potential should a beam of electron be accelerated So that its wavelength becomes equal to $1.54A^{\large\circ}$

$(a)\;63.3V\qquad(b)\;6.33V\qquad(c)\;633V\qquad(d)\;None \;of\; these$

Can you answer this question?

$\lambda = \large\frac{h}{mu}$
$Q.V = \large\frac{1}{2}mu^2$
(where Q is the charge of electron)
$u = \sqrt{\large\frac{2.Q.V}{m}}$
$=\sqrt{\large\frac{2\times1.602\times10^{-19}}{9.108\times10^{-31}}}$
$=5.93\times10^5 \sqrt V$
$\lambda = \large\frac{6.626\times10^{-34}}{9.108\times10^{-31}\times5.93\times10^5\sqrt V}$
$\sqrt V = \large\frac{6.626\times10^{-34}}{9.108\times10^{-31}\times5.93\times10^5\times1.54\times10^{-10}}$
$\;\;\;\;\;\;\;\;\;=7.97$
$\therefore$ V = 63.3 V
Hence anwer is (a)
answered Feb 10, 2014