Question

An unknown particle having double charge as a proton moves with wavelength $\lambda $, if it is accelerated from rest through a potential difference of $\dfrac{V}{8}$ volts. The proton itself moves with the same wavelength $\lambda $ when accelerated from rest through a potential difference ‘V’ volt. The particle is:
A. $H{{e}^{+}}$
B. $L{{i}^{2+}}$ ion
C. $H{{e}^{2+}}$
D. $B{{e}^{2+}}$

Answer 1

Hint: To find out the solution, consider the formula for the de-Broglie wavelength in terms of potential difference and charge for both the particles i.e. the unknown particle and the proton. Then see what is the mass of the particle to find what is the element.

Complete step by step solution:
Given that,
An unknown particle (consider it as X) is having a double charge as proton moves with a wavelength $\lambda $ and when it is accelerated from rest, it has a potential difference $\dfrac{V}{8}$ volts.
The proton is having the same wavelength as the unknown particle as $\lambda $ and it has a potential difference of V volts. We have to find the unknown particle.
So, let us consider de-Broglie wavelength formula, which can be represented as:
$\lambda =\dfrac{h}{\sqrt{2mev}}$ where,
$\lambda $ is the wavelength of the particle,
h is the Planck’s constant,
m is the mass,
e is the charge and
v is the potential difference.
So, let’s the mass of X be ${{M}_{X}}$ and the mass of the proton be ${{M}_{P}}$.
From the question, the charge of X (let it be ${{e}_{X}}$) is double that of protons (let it be $e$).
So, ${{e}_{X}}=2e$. Whereas, the potential difference of the particle (${{V}_{X}}$) is $\dfrac{V}{8}$ volts.
So, the wavelength of the particle (be ${{\lambda }_{X}}$) will be:
${{\lambda }_{X}}=\dfrac{h}{\sqrt{2{{M}_{X}}\times 2e\times \dfrac{V}{8}}}$ ---- (i)
While, the wavelength of proton ($\lambda $) will be:
$\lambda =\dfrac{h}{\sqrt{2{{M}_{P}}eV}}$ ----(ii)
On dividing both, we get:
$\dfrac{{{\lambda }_{X}}}{\lambda }=\dfrac{\sqrt{2{{M}_{P}}eV}}{\sqrt{2{{M}_{X}}\times 2e\times \dfrac{V}{8}}}$
In the question, it is said that the wavelengths of both the particles are equal.
So, $1=\sqrt{\dfrac{4{{M}_{P}}}{{{M}_{X}}}}=2\sqrt{\dfrac{{{M}_{P}}}{{{M}_{X}}}}$
Then, $\dfrac{{{M}_{P}}}{{{M}_{X}}}=\dfrac{1}{4}$
So, ${{M}_{X}}=4{{M}_{P}}$.
The mass of the particle is $4$.
Thus, the particle will be helium which is represented as $He$.
Now, since it has double the charge as that of proton (positive charge). So, the charge upon it will be $2+$.
Therefore, the unknown particle will be $H{{e}^{2+}}$.

Hence, the correct option is C.

Note: de-Broglie’s equation states that a matter acts as a wave such as light and radiation. The equation says about the probability of diffracting a beam of electrons much like a beam of light. It also says that small particles with a mass travelling at a certain speed less than that of light will operate like a wave.