Question

An electron , a proton and an alpha particle have kinetic energies of 16E, 4E and E respectively. What is the qualitative order of their de Broglie wavelengths?

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

Answer 1

The correct option is A ${\lambda }_e>{\lambda }_p={\lambda }_{\alpha }$

As we know,
de Broglie wavelength is
$\lambda =h/mv=h/P=h/\sqrt{2mE}$
so
${\lambda }_e=h/\sqrt{2m_p/1800\times 16E}$
${\lambda }_p=h/\sqrt{2m_p\times 4E}$
${\lambda }_{\alpha }=h/\sqrt{2\times 4m_p\times E}$ so
${\lambda }_e>{\lambda }_p={\lambda }_{\alpha }$