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An electron and a proton, starting from rest, are accelerated through an electric potential difference of...

An electron and a proton, starting from rest, are accelerated through an electric potential difference of the same magnitude. In the process, the electron acquires a speed ve, while the proton acquires a speed vp. Find the ratio ve/vp.

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Expert Solution

Solution:

Since the particles are accelerated from rest, the total energy available in the form of potential energy by the particle is converted to kinetic energy, E = KE

The kinetic energy, K = 1/2 m v²where

m is the mass of the accelerated particle

v is the velocity with which the particle is accelerated

The electrostatic potential energy of the particle is, E = qV where

q is the charge of the particle

V is the potential of the region

For an electron,

$KE= \frac{1}{2}m_{e}v_{e}^{2}$

$E = eV$

Equating these two, $\frac{1}{2} m_{e}v_{e}^{2}= eV$

$v_{e}^{2}=\frac{2 e V }{m_{e}}$

Similarly for a proton,

$\frac{1}{2} m_{p}v_{p}^{2}= qV$

$v_{p}^{2}=\frac{2qV}{m_{p}}$

Taking the ratio of v_e and v_p

$\frac{v_{e}^{2}}{v_{p}^{2}} = \frac{e\times m_{p}}{q\times m_{e}}$ ( cancelling potential from numerator and denominator since the potential experienced is the same for both the particles)

$\frac{v_{e}}{v_{p}} = \sqrt{\frac{e\times m_{p}}{q\times m_{e}}}$

since the magnitude of charges are the same we can cancel out that also

Therefore, $\frac{v_{e}}{v_{p}} = \sqrt{\frac{ m_{p}}{ m_{e}}} =\sqrt{\frac{1.67\times 10^{-27}}{9.1\times 10^{-31}}} = 42.83$

genius_generous answered 1 month ago

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