Question

If $E_1,E_2$ and $E_3$ respectively represent the kinetic energies of an electron, an $\alpha$-particle and a proton each having same de-Broglie’s wavelength, then

• A. $E_1>E_2>E_3$
• B. $E_2>E_3>E_1$
• C. $E_1>E_3>E_2$
• D. $E_1=E_2=E_3$

Answer 1

The correct option is C $E_1>E_3>E_2$

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

Since $p=\frac{h}{\lambda }$ for particle of de-Broglie wavelength $\lambda$, we can say
$E=\frac{h^2}{2m{\lambda }^2}$

Given, ${\lambda }_1={\lambda }_2={\lambda }_3$
$\Rightarrow E\propto \frac{1}{m}$

Since $m_{\alpha }>m_p>m_e$ i.e $m_2>m_3>m_1$, we get
$E_1>E_3>E_2$