Question

An electron and proton are accelerated through the same potential difference. The ratio of their de-Broglie wavelengths will be
[ Taking mp and me as mass' of proton and mass of electron ]

• A. ${\left(\frac{m_p}{m_e}\right)}^{3/2}$
• B. ${\left(\frac{m_p}{m_e}\right)}^{1/2}$
• C. ${\left(\frac{m_p}{m_e}\right)}^{5/2}$
• D. $1$

Answer 1

Answer: (B)

${\left(\frac{m_p}{m_e}\right)}^{1/2}$

$\lambda =\frac{h}{\sqrt{2mE}}$
$\Rightarrow {\frac{{\lambda }_e}{\lambda }}_p={\left(\frac{m_p}{m_e}\right)}^{\frac{1}{2}}$