Question

A particle carrying 5 electrons starts from rest and is accelerated through a potential difference of 9000V. Calculate the K.E.acquired by it in MeV.(charge on electron = 1.6 * ${10}^{-}19$C)

• A. $\frac{1.78\times {10}^{-10}}{{10}^6}MeV$
• B. $\frac{1.78\times {10}^{-13}}{{10}^1}MeV$
• C. $\frac{1.78\times {10}^{-13}}{{10}^6}MeV$
• D. 0.90 MeV

Answer 1

The correct option is C $\frac{1.78\times {10}^{-13}}{{10}^6}MeV$

$\begin{array}{c}Accoring\\ KE=qV\\ =2\times 1.6\times {10}^{-19}\times 2000=4000eV\left(asalphaparticlehas2eche\right)\\ letcheofneucleasisQ\\ then\\ 4000eV=\frac{2eQ}{4\pi {\epsilon }_0\times 8000A^0}\\ Q=\frac{4000\times 8000\times {10}^{-10}\times {10}^{-9}}{9\times 2}C\\ =\frac{1.78\times {10}^{-13}}{{10}^6}MeV\end{array}$

Hence, the option $C$ is the correct answer.