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 × 107m/s  (B)  text1 text.9 ! !× ! ! text 1 text06 textm/s  (C)  text1 text.9 ! !× ! ! text 1 text07 textm/s  (D)  text5 text.7 ! !× ! ! text 1 text07 textm/s

Tardigrade Admin · Accepted Answer

If the voltage given is V , then the energy of electron =(1/2)mv2=eV v=√(2eV/m) =√(2× 1.6× 10-19× 1000/9.1× 10-31) =1.875× 107 =1.9× 107 m/s

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

$3.8 \times 10^{7} m / s$

$1 .9 \times 1 0^{6} m/s$

$1 .9 \times 1 0^{7} m/s$

$5 .7 \times 1 0^{7} m/s$

If the voltage given is $V$ , then the energy of electron $= \frac{1}{2} m v^{2} = e V$ $v = \frac{2 e V}{m}$ $= \frac{2 \times 1.6 \times 10 ^{- 19} \times 1000}{9.1 \times 10 ^{- 31}}$ $= 1.875 \times 10^{7}$ $= 1.9 \times 10^{7} m / s$

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

3.8×107m/s

1.9 × 106m/s

1.9 × 107m/s

5.7 × 107m/s

If the voltage given is V , then the energy of electron =21​mv2=eV v=m2eV​​ =9.1×10−312×1.6×10−19×1000​​ =1.875×107 =1.9×107m/s

$3.8 \times 10^{7} m / s$

$1 .9 \times 1 0^{6} m/s$

$1 .9 \times 1 0^{7} m/s$

$5 .7 \times 1 0^{7} m/s$

If the voltage given is $V$ , then the energy of electron $= \frac{1}{2} m v^{2} = e V$ $v = \frac{2 e V}{m}$ $= \frac{2 \times 1.6 \times 10 ^{- 19} \times 1000}{9.1 \times 10 ^{- 31}}$ $= 1.875 \times 10^{7}$ $= 1.9 \times 10^{7} m / s$