Question

The velocity of electrons accelerated by potential difference of $1\times {10}^{4}V$ (The charge of the elctron is $1.6\times {10}^{-19}C$ and mass is $9.11\times {10}^{-31}$kg) is

• A. $5.93\times {10}^{7}m{s}^{-1}$
• B. $2.94\times {10}^{7}m{s}^{-1}$
• C. $6.87\times {10}^{7}m{s}^{-1}$
• D. $3.98\times {10}^{7}m{s}^{-1}$

Answer 1

The correct option is A $5.93\times {10}^{7}m{s}^{-1}$

K.E. = eV
$\frac{1}{2}m{V}^2=eV$

$V^2=\frac{2eV}{m}$

$V=\sqrt{\frac{2eV}{m}}$

= $\sqrt{\frac{2\times 1.6\times {10}^{-19}\times {10}^4}{9.11\times {10}^{-31}}}$

= $0.593\times {10}^8$

= 5.93 x ${10}^7\frac{m}{s}$.

Answer is A.