Question

If K.E. of an electron, $\alpha -particle$ and a proton are represented by $E_1,E_2$ and $E_3$ respectively (provided all these have same de Broglie wavelength) then______________.

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

Answer 1

Answer: (C)

$E_1>E_3>E_2$

$KE=\frac{1}{2}m{u}^2$ and $\lambda =\frac{h}{mu}$

$KE=\frac{1}{2}\cdot \frac{h^2}{m{\lambda }^2}=\frac{h^2}{2m{\lambda }^2}$

Given, all these have the same de Broglie wavelength

$E_1=\frac{h^2}{2m_e{\lambda }^2}$, $E_2=\frac{h^2}{2m_p{\lambda }^2}$

$E_3=\frac{h^2}{2m_{\alpha }{\lambda }^2}$
$E\propto \frac{1}{m}$

$m_e<m_{\alpha }<m_p$

So lesser is the mass, higher is the kinetic energy so the order is

$E_1>E_3>E_2$