An electron moving in a electric potential field $V_{1}$ enters a higher electric potential field $V_{2}$ then the change in kinetic energy of the electron is proportional to

Open in App

Solution

The higher potential is given as $V_{2}$

The lower potential is given as $V_{1}$

Hence, the potential difference is calculated as $(V_{2} - V_{1})$

The kinetic energy gained by the electron due to the potential difference is the product of the charge of electron with the potential difference.

Mathematically kinetic energy K.E $= e (V_{2} - V_{1})$ [1]

We are asked to calculate the velocity of electron.

Let the velocity of the electron is $v$ and mass is $m$

Hence, the kinetic energy of the electron is calculated as -

$K . E = \frac{1}{2} m v^{2}$ [2]

From 1 and 2 we get,

$⟹ \frac{1}{2} m v^{2} = e (V_{2} - V_{1})$

$⟹ v^{2} = \frac{2 e}{m} V_{2} - V_{1})$

$⟹ v = \sqrt{\frac{2 e}{m} V_{2} - V_{1})}$

Suggest Corrections

Similar questions

6.4 \times 10^{- 19} J

ϕ_{1}, ϕ_{2}

(V_{C} - V_{A})

A

B

1.28 \times 10^{- 18} J

ϕ_{1}, ϕ_{2}

V_{E} - V_{D}