A proton and a \(\mathrm{Li}^{3+}\) nucleus are accelerated by the same potential. If \(\lambda_{\mathrm{Li}}\) and \(\lambda_{\mathrm{p}}\) denote the de Broglie wavelengths of \(\mathrm{Li}^{3+}\) and proton respectively, then the value of \(\frac{\lambda_{\mathrm{Li}}}{\lambda_{\mathrm{p}}}\) is \(x \times 10^{-1}\). The value of \(x\) is ____. [Rounded off to the nearest integer] [Mass of \(\mathrm{Li}^{3+}\) = 8.3 × mass of proton]

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Atomic Structure

A proton and a

Li^{3 +}

nucleus are accelerated by the same potential. If

λ_{Li}

and

λ_{p}

denote the de Broglie wavelengths of

Li^{3 +}

and proton respectively, then the value of

\frac{λ _{Li}}{λ _{p}}

x \times 1 0^{- 1}

. The value of

x

is ____. [Rounded off to the nearest integer]
[Mass of

Li^{3 +}

= 8.3 × mass of proton]

Answer: 2

The de Broglie wavelength is given by:

λ = \frac{h}{p} = \frac{h}{2 m K}

where K is the kinetic energy.

When a charged particle is accelerated through a potential V:

K = q V

For the proton:

K_{p} = e V

(charge = e)

For

Li^{3 +}

K_{Li} = 3 e V

(charge = 3e)

Ratio of wavelengths:

\frac{λ _{Li}}{λ _{p}} = \frac{2 m _{p} \cdot e V}{2 m _{Li} \cdot 3 e V}

= \frac{m _{p}}{3 m _{Li}}

Given

m_{Li} = 8.3 m_{p}

\frac{λ _{Li}}{λ _{p}} = \frac{m _{p}}{3 \times 8.3 m _{p}} = \frac{1}{24.9} \approx \frac{1}{25} = \frac{1}{5} = 0.2 = 2 \times 1 0^{- 1}

Therefore,

x = 2

Answer: 2

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Integer answer