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

Want to ask us a question? Click here

Browse Questions

Ad |

0 votes

0 votes

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, $p = mv$

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

$= 4.04 \times 10^{-24} kg\: m/s$

Ask Question

Tag:MathPhyChemBioOther

Take Test

x

JEE MAIN, CBSE, NEET Mobile and Tablet App

The ultimate mobile app to help you crack your examinations

...