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

$\frac{\nu_{1} - \nu_{2}}{k - 1}$

$1.9 \times 10^{6}m/s$

$1.9 \times 10^{7}m/s$

$2\sqrt{2}V$

Answer

$1.9 \times 10^{7}m/s$

Explanation

If the voltage given is V, then the energy of electron

$\frac{1}{2}mv^{2} = eV \Rightarrow v = \sqrt{\frac{2eV}{m}}$

$= \sqrt{\frac{2 \times 1.6 \times 10^{- 19} \times 1000}{9.1 \times 10^{- 31}}} = 1.875 \times 10^{7} \approx 1.9 \times 10^{7}m/s$

Question: An electron is accelerated through a potential difference of 1000 volts. Its velocity is nearly....