Question

A neutron, one of the constituents of a nucleus, is found to pass two points 60 metres apart in a time interval of $1.8\times {10}^{-4}sec$. The mass of the neutron is $1.67\times {10}^{-27}kg$. Assuming that the speed is constant, its kinetic energy is

• A. $9.3\times {10}^{-17}joule$
• B. $9.3\times {10}^{-14}joule$
• C. $9.3\times {10}^{-21}joule$
• D. $9.3\times {10}^{-11}joule$

Answer 1

The correct option is A $9.3\times {10}^{-17}joule$

$\begin{array}{c}\text{Speed}=\frac{\text{Distance}}{\text{time}}\\ v=\frac{60}{18\times {10}^{-5}}\\ v=\frac{{10}^6}{3}m/s\\ m=1.67\times {10}^{-27}kg\\ \text{Kinetic energy,}k=\frac{1}{2}m{v}^2\end{array}$

$\begin{array}{cc}k& =\left(\frac{1}{2}\right)(1.67\times {10}^{-27}){\left(\frac{106}{3}\right)}^2\\ & k=\left(\frac{1}{2}\right)(1.67\times {10}^{-27})\frac{{10}^{12}}{9}\\ & k=9.3\times {10}^{-17}\text{Joules}\end{array}$