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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 binding energy of electron in a metal is 250 KJ/mol. The threshold frequency of metal is


1 Answer

$hr_{\large\circ} = 250\times10^3J/mol$
$\therefore r_{\large\circ} = \large\frac{250\times10^3}{6.626\times10^{-34}\times6.023\times10^{23}}$
Hence answer is (b)
answered Feb 10, 2014 by sharmaaparna1

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