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# The momentum of the electrons accelerated through a potential difference of $56\; V$ is

$\begin {array} {1 1} (a)\;0.404 \times 10^{-25} kg\: m/s & \quad (b)\; 40.4 \times 10^{-24} kg \: m/s \\ (c)\; 4.04 \times 10^{-22}kg \: m/s & \quad (d)\;4.04 \times 10^{-24} kg\: m/s \end {array}$

Can you answer this question?

And : (d)
At equilibrium, the kinetic energy of each electron is equal to accelerating potential,
i.e, $\large\frac{1}{2}\: mv^2\: = eV$
$\Rightarrow v^2 = 2eV/m$
So, $v=\sqrt{\bigg[\large\frac{(2 \times 1.6 \times 10^{-19} \times 56)}{(9.1 \times 10^{-31} )}\bigg]}$
$= \sqrt{[19.69 \times 10^{12} ]} = 4.44 \times 10^6 \: m/s$
So, momentum, $p = mv$
$= 9.1 \times 10^{-31} \times 4.44 \times 10^6$
$= 4.04 \times 10^{-24} kg\: m/s$

answered Dec 25, 2013
edited Mar 13, 2014