An electron is accelerated through a potential difference of 1000 V, its velocity is nearly

Question

An electron is accelerated through a potential difference of 1000 V, its velocity is nearly (A) 3.8 × 107 ms -1 (B) 1.9 × 106 ms -1 (C) 1.9 × 107 ms -1 (D) 5.7 × 107 ms -1

Tardigrade Admin · Accepted Answer

When an electron is accelerated by a potential difference of

V

volts, then
or

e V = U

This potential energy is converted into

K E

of electron,

K E = U

\Rightarrow \frac{1}{2} m v^{2} = e V

\Rightarrow v = \frac{2 e V}{m} = \frac{2 \times 1.6 \times 1 0 ^{- 19} \times 1000}{9.1 \times 1 0 ^{- 31}}

= 1.875 \times 1 0^{7}

\approx 1.9 \times 1 0^{7} m s^{- 1}

Q. An electron is accelerated through a potential difference of $1000 V$ , its velocity is nearly

$3.8 \times 1 0^{7} m s^{- 1}$

$1.9 \times 1 0^{6} m s^{- 1}$

$1.9 \times 1 0^{7} m s^{- 1}$

$5.7 \times 1 0^{7} m s^{- 1}$

Q. An electron is accelerated through a potential difference of 1000V, its velocity is nearly

3.8×107ms−1

1.9×106ms−1

1.9×107ms−1

5.7×107ms−1

When an electron is accelerated by a potential difference of V volts, then or eV=U This potential energy is converted into KE of electron, KE=U ⇒21​mv2=eV ⇒v=m2eV​​=9.1×10−312×1.6×10−19×1000​​ =1.875×107 ≈1.9×107ms−1

Q. An electron is accelerated through a potential difference of $1000 V$ , its velocity is nearly

$3.8 \times 1 0^{7} m s^{- 1}$

$1.9 \times 1 0^{6} m s^{- 1}$

$1.9 \times 1 0^{7} m s^{- 1}$

$5.7 \times 1 0^{7} m s^{- 1}$