A proton and a Li^3+ nucleus are accelerated by the same potential. If \lambda _Li and \lambda _p denote the de Broglie wavelengths of Li^3+ and proton respectively, then the value of \lambda _Li \over \lambda _p is x \times 10-1. The value of x is \underline\hspace2cm . (Rounded off to the nearest integer) [Mass of Li^3+ = 8.3 mass of proton]

Structure of Atom PYQ — JEE Main 2021 · 24 Feb · Shift 1 · Q16 with Solution

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Correct answer: 2

For a particle,

\lambda = \frac{h}{p}

where $p$ is momentum.

If a charged particle is accelerated through a potential difference $V$ , then its kinetic energy is

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

So,

p = \sqrt{2mqV}

Hence,

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

Thus,

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

for the same accelerating potential $V$ .

For proton:

For $Li^{3+}$ nucleus:

Therefore,

\frac{\lambda_{Li}}{\lambda_p} = \sqrt{\frac{m_p\,e}{(8.3m_p)(3e)}}

Cancel $m_p$ and $e$ :

\frac{\lambda_{Li}}{\lambda_p} = \frac{1}{\sqrt{8.3\times 3}}

8.3 \times 3 = 24.9

So,

\frac{\lambda_{Li}}{\lambda_p} = \frac{1}{\sqrt{24.9}}

Now,

\sqrt{24.9} \approx 4.99

Thus,

\frac{\lambda_{Li}}{\lambda_p} \approx \frac{1}{4.99} \approx 0.200

Given,

\frac{\lambda_{Li}}{\lambda_p} = x \times 10^{-1}

Since,

0.200 = 2.00 \times 10^{-1}

So,

x = 2

Derived answer: $2$

Stored correct answer: $2$

Both match.