Question

Calculate the

(a) Momentum

(b) De Broglie wavelength of the electrons accelerated through a potential difference of 56 V.

Answer 1

(a) Potential difference, $V=56volts$

At equilibrium, Kinetic energy is accelerating potential.

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

$v=4.44\times {10}^{6}m/s$

$P=mv$

$\Rightarrow 9.1\times {10}^{-31}\times 4.44\times {10}^6=4.04\times {10}^{-24}$

where $m$ is the mass of the electron.

(b) De Broglie wavelength is given by

$\lambda =\frac{12.27\times {10}^{-10}}{\sqrt{V}}$

$\Rightarrow \frac{12.27\times {10}^{-10}}{\sqrt{56}}=0.164nm$