Question

Calculate the de Broglie wavelength for a beam of electron whose energy is 100ev:
A. $1{{A}^{{}^\circ }}$
B. $1.23{{A}^{{}^\circ }}$
C. $2.46{{A}^{{}^\circ }}$
D. None of these

Answer 1

De Broglie wavelength is an important concept of quantum mechanics. The wavelength represented by $\lambda$which is associated with an object in relation to its momentum and mass and known as de Broglie wavelength.

Complete step by step solution:
De Broglie gives the concept that matter can show dual nature it can act as wave and particles as well for example light as light can behave both as a wave i.e. it can be diffracted and has a wavelength and also act as a particles it as it contains packets of energy represented by $h\nu$.
De Broglie wavelength is given by the formula
$\lambda =\dfrac{h}{p}$; where h = planck’s constant and p is momentum which is given by $p=mv$
Hence $\lambda$can be written as $\lambda =\dfrac{h}{mv}$
As we know that
$E=\dfrac{1}{2}m{{v}^{2}}=100ev=100\times 1.6\times {{10}^{-19}}J$
$\implies {{v}^{2}}=\dfrac{2E}{m}$
$\therefore v={{(\dfrac{2E}{m})}^{\dfrac{1}{2}}}$
$\lambda =\dfrac{h}{mv}$
Putting the value of v
$\lambda =\dfrac{h}{\sqrt{2mE}}$
$\lambda =\dfrac{6.6\times {{10}^{-34}}}{\sqrt{2\times 9.1\times {{10}^{-31}}\times 100\times 1.6\times {{10}^{-19}}}}$
$\implies$ $1.23\times {{10}^{-10}}$m
$\therefore$ $1.23{{A}^{{}^\circ }}$

Hence option B is the correct answer.

Note: The wave properties of matter are only observable for very small objects in which the de Broglie wavelength of a double-slit interference pattern is produced by using electrons as the source. That’s why a crystal acts as a diffraction grating for electrons.