An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

Question

An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is (A) 2.0 × 108 ms-1 (B) 1.88 × 107 ms-1 (C) 1.5 × 107 ms-1 (D) 1.5 × 108 ms-1

Tardigrade Admin · Accepted Answer

K.E. acquired by electron. (1/2)mv2 = eV or v =√(2eV/m) i.e. v = √(2 × 1.6 × 10-19 × 1000/9.1 × 10-31) = 1.88 × 107 m s-1

Q. An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

$2.0 \times 1 0^{8} m s^{- 1}$

$1.88 \times 1 0^{7} m s^{- 1}$

$1.5 \times 1 0^{7} m s^{- 1}$

$1.5 \times 1 0^{8} m s^{- 1}$

K.E. acquired by electron. $\frac{1}{2} m v^{2}$ = eV or $v = \frac{2 e V}{m}$ i.e. $v = \frac{2 \times 1.6 \times 1 0 ^{- 19} \times 1000}{9.1 \times 1 0 ^{- 31}}$ = 1.88 $\times 1 0^{7} m s^{- 1}$

Q. An electron is accelerated through a potential difference of 1000 volt. The velocity acquired by the electron is

2.0×108ms−1

1.88×107ms−1

1.5×107ms−1

1.5×108ms−1

K.E. acquired by electron. 21​mv2 = eV or v=m2eV​​ i.e. v=9.1×10−312×1.6×10−19×1000​​ = 1.88 ×107ms−1

$2.0 \times 1 0^{8} m s^{- 1}$

$1.88 \times 1 0^{7} m s^{- 1}$

$1.5 \times 1 0^{7} m s^{- 1}$

$1.5 \times 1 0^{8} m s^{- 1}$

K.E. acquired by electron. $\frac{1}{2} m v^{2}$ = eV or $v = \frac{2 e V}{m}$ i.e. $v = \frac{2 \times 1.6 \times 1 0 ^{- 19} \times 1000}{9.1 \times 1 0 ^{- 31}}$ = 1.88 $\times 1 0^{7} m s^{- 1}$