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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


1 Answer

$\lambda = \large\frac{h}{mu}$----(1)
$u = \sqrt{\large\frac{3RT}{m}}$
=1367.8 m/Sec
Now substitute the value of u in(1)
Hence answer is (a)
answered Feb 9, 2014 by sharmaaparna1

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