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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 velocity must an electron travel so that its momentum is equal to that of photon of wavelength of $\lambda = 5200A^{\large\circ}$

$(a)\;800m/s\qquad(b)\;1400m/s\qquad(c)\;400m/s\qquad(d)\;200m/s$

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Momentum mu = momentum of wavelength $5200A^{\large\circ}$
$= 9.108\times10^{-31}\times u$
$= \large\frac{6.626\times10^{-34}}{5200\times10^{-10}}$
$mu = \large\frac{h}{\lambda}$
$\therefore 4 = 1400 m sec^{-1}$
Hence answer is (b)
answered Feb 9, 2014 by sharmaaparna1
 

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