Question

De Broglie wavelengths of a proton are $\lambda$ at ${27}^{0}C$. At ${927}^{0}C$, the De Broglie wavelength would be :

• A. $\lambda$
• B. $\frac{\lambda }{2}$
• C. $\frac{\lambda }{3}$
• D. $\frac{\lambda }{4}$

Answer 1

The correct option is B $\frac{\lambda }{2}$

According to de-Broglie,
$\lambda =\frac{h}{mu}=\frac{h}{\sqrt{2m(K.E.)}}=\frac{h}{\sqrt{2m\times k^{{}^{\prime }}\times T}}$ $(KE\propto T)$
$\therefore \lambda \propto \sqrt{\frac{1}{T}}$
or $\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{T_2}{T_1}}$
or $\frac{\lambda }{{\lambda }_2}=\sqrt{\frac{1200}{300}}$
$\therefore {\lambda }_2=\frac{\lambda }{2}$