An electron and a proton starting from rest get accelerated through po

An electron and a proton starting from rest get accelerated through potential difference of 100 kV. The final speeds of the electron and the proton are V$_e$ and V$_p$ respectively. Which one of the following relations is correct?

”V<tex>$_e$</tex>

”Cannot

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Answer is Wrong!

This question was previously asked in

UPSC CDS-1 – 2019

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The correct option is A.

When a charged particle is accelerated from rest through a potential difference V, its kinetic energy is equal to the work done on it by the electric field, which is |qV|. So, (1/2)mv^2 = |qV|. The speed is given by v = sqrt(2|q|V/m). The electron and the proton have the same magnitude of charge (|q| = e) and are accelerated through the same potential difference V. Thus, their kinetic energies gained are equal: KE_e = KE_p = eV.
(1/2)m_e * V_e^2 = (1/2)m_p * V_p^2
m_e * V_e^2 = m_p * V_p^2
V_e / V_p = sqrt(m_p / m_e)
Since the mass of the electron (m_e) is significantly less than the mass of the proton (m_p), the ratio m_p / m_e is greater than 1. Therefore, V_e / V_p > 1, which means V_e > V_p. The lighter particle (electron) achieves a much higher speed than the heavier particle (proton) for the same kinetic energy.

The mass of an electron is approximately 9.11 x 10^-31 kg, and the mass of a proton is approximately 1.67 x 10^-27 kg. The proton mass is about 1836 times the electron mass. Thus, the electron’s speed will be about sqrt(1836) ≈ 43 times the proton’s speed.

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