The de-Broglie wavelength associated with an electron and a proton were calculated by accelerating them through same potential of 100 V. What should nearly be the ratio of their wavelengths? (mp = 1.00727u me = 0.00055u)

Dual Nature of Radiation PYQ — JEE Main 2021 · 16 Mar · Shift 2 · Q53 with Solution

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Correct answer: D

For a particle accelerated through a potential difference $V$ ,

\lambda = \frac{h}{p}

and the gained kinetic energy is

qV = \frac{p^2}{2m}

So,

p = \sqrt{2mqV}

Hence,

\lambda = \frac{h}{\sqrt{2mqV}}

For both electron and proton, $V$ is same and $q=e$ is also same. Therefore,

\lambda \propto \frac{1}{\sqrt{m}}

So the ratio is

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{m_p}{m_e}}

Given:

m_p = 1.00727\,u, \qquad m_e = 0.00055\,u

Thus,

\frac{\lambda_e}{\lambda_p} = \sqrt{\frac{1.00727}{0.00055}}

First calculate the mass ratio:

\frac{1.00727}{0.00055} \approx 1831.4

Now take square root:

\sqrt{1831.4} \approx 42.8

So,

\lambda_e : \lambda_p \approx 42.8 : 1 \approx 43 : 1

Therefore, the correct answer is:

\boxed{43:1}