Question

Calculate the speed of a proton that is accelerated from rest through an electric potential difference of 163 v. (b) calculate the speed of an electron that is accelerated through the same potential difference.

Answer 1

We could use the energy conservation:

      Kinetic energy = Potential energy
=> $\frac{1}{2}$ M $v^{2}$   =  eV
where, M = mass of proton
            v = speed
            e = charge 
            V = potential difference through which it is accelerated

So, finding speed using the above equation:

v = $\sqrt{\frac{2eV}{M}}$

Putting values,
v =  $\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 163}{1.67 \times 10^{-27}}}$
v = 177.15 Km/s

Now, let  m  = mass of electron

So, Above we got the formula for the speed of proton accelerated through V.

For electron, just replace the mass M (proton) with m (electron) and that's it. Because V is same for both.

So, Speed of electron, v =  $\sqrt{\frac{2eV}{m}}$

Putting values.

v =  $\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 163}{9.1 \times 10^{-31}}}$

On solving
v = 7.56 x $10^{6}$ m/s   or  7560 km/s

Answer 2

You are given the electric potential difference of a proton at 163V. You are asked to find the speed of a proton from rest and when moving but the same potential difference. Use the energy of photon equation.

from rest with potential difference of 163 V
E = 1/2 mv² = eΔV
v = √$\sqrt{ \frac{2e(deltaV)}{ m_{p} } } = \sqrt{ \frac{2 * 1.6 x 10^{-19} *163 }{1.67 x 10^{-27} } }$
v = 176,730 m/s or 176.73 km/s

electron that is accelerated through the same potential difference
E = 1/2 mv² = eΔV
v = √$\sqrt{ \frac{2e(deltaV)}{ m_{e} } } = \sqrt{ \frac{2 * 1.6 x 10^{-19} *163 }{9.11 x 10^{-31} } }$
v = 7566753 m/s = 7,566.8 km/s