Question

An electron and a proton possess the same amount of Kinetic energy. Then the relation between the wavelength of electron$\left({\lambda }_e\right)$ and the wavelength of proton$\left({\lambda }_p\right)$ is

• A. ${\lambda }_e={\lambda }_p$
• B. ${\lambda }_e>{\lambda }_p$
• C. ${\lambda }_e<{\lambda }_p$
• D. ${\lambda }_e\le {\lambda }_p$

Answer 1

Answer: (B)

For an electron
$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2Em}$

${\lambda }_e=\frac{h}{mv}$

$=\frac{h}{\sqrt{2E{m}_e}}$

For a proton

$\frac{1}{2}m{v}^2=E$

$mv=\sqrt{2Em}$

${\lambda }_p=\frac{h}{\sqrt{2E{m}_p}}$

$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}$
$\frac{{\lambda }_e}{{\lambda }_p}>1$
${\lambda }_e>{\lambda }_p$