Question

If kinetic energy of an electron is reduced by $\left(1/9\right)$ then how many times its de Broglie wavelength will increase?

• A. $3$
• B. $4$
• C. $5$
• D. $6$

Answer 1

The correct option is A $3$

K.E of electronreduced by $\frac{1}{9}$

K.E=$\frac{p^2}{2m}$$=\frac{h^2}{{2m\lambda }^2}$

$\Rightarrow =$$K.E\propto$${\frac{1}{\lambda }}^2$

[$\because$debroglie wave length,$\lambda =\frac{h}{p}$]

$\frac{K.E_1}{{K.E}_2}=\frac{{\lambda }_2^2}{{\lambda }^2}$

$\frac{1}{\left(V_9\right)}={\left(\frac{{\lambda }_2}{{\lambda }_1}\right)}^2\Rightarrow {\lambda }_2=3{\lambda }_1$