Question

If an electron is accelerated from rest through a potential difference of 9.9 kV, what is its resulting speed

Answer 1

Answer:

v = 5.9 x 10⁷ m/s

Explanation:

The kinetic energy of the electron in terms of potential difference is given as:

$K.E = eV$--------------- equation (1)

where,

e = charge on electron = 1.6 x 10⁻¹⁹ C

V = Potential Difference = 9.9 KV = 9900 Volts

The kinetic energy in general is given as:

$K.E = \frac{1}{2}mv^{2}$--------- equation (2)

where,

m = mass of electron = 9.1 x 10⁻³¹ kg

v = speed of electron = ?

Therefore, comparing equation (1) and equation (2), we get:

$\\\frac{1}{2}mv^{2} = eV\\\\\frac{1}{2}(9.1\ x\ 10^{-31}\ kg)v^{2} = (1.6\ x\ 10^{-19}\ C)(9900\ volts)\\\\v = \sqrt{34.81\ x\ 10^{14}}$

v = 5.9 x 10⁷ m/s