Question

de Broglie wavelength for a beam of electron having energy 100 eV in $A^0$ is :

• A. $1.23$
• B. $2.46$
• C. $1.48$
• D. $2.45$

Answer 1

The correct option is A $1.23$

E = $\frac{1}{2}$mv${}^2$ = 100 eV = 100 $\times$ 1.6 $\times$ 10${}^{-19}$ J
v${}^2$ = $\frac{2E}{m}$
v = ($\frac{2E}{m}{)}^{\left(1/2\right)}$ = $\sqrt{2mE}$
$\lambda$ = $\frac{h}{mv}$ = $\frac{h}{\sqrt{2mE}}$ metre
$\lambda$ = ${\frac{6.6\times {10}^{-34}}{(2\times 9.1\times {10}^{-31}\times 100\times 1.6\times {10}^{-19}}}^{1/2}$
=1.23 $\times$ 10${}^{-10}$ m = $1.23A^0$