Question

The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of $100\text{V}$. What should nearly be the ratio of their wavelengths? $(m_p=1.00727u,m_e=0.00055u)$

• A. $(1860{)}^2:1$
• B. $43:1$
• C. $1860:1$
• D. $41.4:1$

Answer 1

The correct option is B $43:1$

De-broglie wavelength is given as: $$\lambda =\frac{h}{\sqrt{2mqV}}$$ Since magnitude of charge $\left(q\right)$ and potential difference $\left(V\right)$ through which they are accelerated is same.

$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{\frac{m_p}{m_e}}=\sqrt{\frac{1.00727}{0.00055}}$
$\frac{{\lambda }_e}{{\lambda }_p}=\sqrt{1831}\simeq 43$