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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 . The wavelength of helium atom whose speed is equal to its rms speed at $27^{\large\circ}$ C

$(a)\;7.29\times10^{-11}m\qquad(b)\;4.28\times10^{-10}m\qquad(c)\;5.31\times10^{-11}m\qquad(d)\;6.28\times10^{-11}m$

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A)
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$\lambda = \large\frac{h}{mu}$----(1)
$u = \sqrt{\large\frac{3RT}{m}}$
$=\sqrt{\large\frac{3\times8.314\times300}{4\times10^{-3}}}$
=1367.8 m/Sec
Now substitute the value of u in(1)
$=\large\frac{6.626\times10^{-34}}{\large\frac{4\times10^{-3}\times1367.8}{6.023\times10^{23}}}$
$=7.29\times10^{-11}m$
Hence answer is (a)
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