Question

An electron, $\alpha$- particle and a proton have the same kinetic energy. A comparison of their de Broglie wavelengths yields:

• A. ${\lambda }_p>{\lambda }_a<{\lambda }_c$
• B. ${\lambda }_e>{\lambda }_a<{\lambda }_p$
• C. ${\lambda }_a<{\lambda }_p<{\lambda }_e$
• D. ${\lambda }_p<{\lambda }_e<{\lambda }_a$

Answer 1

The correct option is C ${\lambda }_a<{\lambda }_p<{\lambda }_e$

$e^{-},\alpha ,p$

${\lambda }_{debroglie}=\frac{h}{mv}=\frac{h}{p}$

all have same $KE$

$KE=\frac{p^2}{2m}$ $m_{\alpha }>m_p>m_e$

$p=\sqrt{2mk}$ So ${\lambda }_{e^{-}}>{\lambda }_p>{\lambda }_{\alpha }$

$p_e=\sqrt{2m_{e}k}$

$p_p=\sqrt{2m_{p}k}$

$p_{\alpha }=\sqrt{2m_{\alpha }k}$

Thus, wavelength is inversely propotional to the mass. Thus, the wavelength of electron will be maximum as it has a smallest mass.