Ask Questions, Get Answers

Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

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. If uncertainities in position and momentum of an electron are same , then uncertainity in its velocity can be given by

$(a)\;\geq\sqrt{\large\frac{h}{4\pi m^2}}\qquad(b)\;\geq\sqrt{\large\frac{4\lambda}{4\pi m}}\qquad(c)\;\geq\sqrt{\large\frac{h}{2m^2}}\qquad(d)\;Either\; of\; these$

1 Answer

$\bigtriangleup u = \bigtriangleup p$
$\therefore (\bigtriangleup p)^2 \geq \large\frac{h}{4\pi}$
$(\bigtriangleup um)^2 \geq \large\frac{h}{4\pi}$
$\therefore \bigtriangleup u \geq \sqrt{\large\frac{h}{4\pi m^2}} \geq \sqrt{\large\frac{4\lambda}{4\pi m}}\geq \sqrt{\large\frac{h}{2m^2}}$
Since $\sqrt h = \large\frac{h}{2\pi} and \lambda = \large\frac{h}{mu}$
Hence the anwer is (d)
answered Feb 10, 2014 by sharmaaparna1

Related questions