Question

An electron and a proton both are accelerated through the same potential, then the ration $\dfrac{{{\lambda }_{e}}}{{{\lambda }_{p}}}$will be?
A- 1
B- $\dfrac{{{m}_{e}}}{{{m}_{p}}}$
C- $\dfrac{{{m}_{p}}}{{{m}_{e}}}$
D- $\sqrt{\dfrac{{{m}_{p}}}{{{m}_{e}}}}$

Answer 1

we know whenever any charged particle is accelerated through some region having potential difference then we can always write the relationship between the debroglie wavelength emitted by that charged particle and the value of its kinetic energy or in other terms the accelerating potential difference

Complete step by step answer:
From De Broglie wavelength relationship, the relation between the value of the wavelength of the wave emitted by the charged particle and the accelerating potential difference is given by $\lambda =\dfrac{h}{\sqrt{2mqV}}$, where q is the charge of the particle, m is its mass, V is the accelerating potential.
Here we are given two charged particles one is proton and the other is electron, since the magnitude of the charge on both of them is same, thus,
$\Rightarrow\dfrac{{{\lambda}_{e}}}{{{\lambda}_{p}}}=\dfrac{\dfrac{h}{\sqrt{2{{m}_{e}}qV}}}{\dfrac{h}{\sqrt{2{{m}_{p}}qV}}} \\\ \therefore \dfrac{{{\lambda }_{e}}}{{{\lambda }_{p}}}=\sqrt{\dfrac{{{m}_{p}}}{{{m}_{e}}}} \\\$

So, the correct answer is “Option D”.

Additional Information:
According to de broglie wavelength hypothesis, whenever a body having some mass moves with speed then waves get emitted by it and these are called matter waves and the wavelength of such is dependent on the mass and the velocity of the body.

Note:
We can also write the above-mentioned relationship in terms of kinetic energy where we know when some charged particle is accelerated through a potential difference, V and its kinetic energy is given by qV. Also, the mass of the proton is greater than the mass of the electron. The charge on proton and the charge on electron both are same in magnitude but opposite in polarity.