Question

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

Answer 1

v = 5.9 x 10⁷ m/s

Step-by-step 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 = (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:

`\\(1)/(2)mv^(2) = eV\\\\(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