Question

Determine:

a. The speed of a $7.0MeV$neutron.

b. The speed of a $15MeV$helium atom.

c. The specific type of particle that has $1.14keV$of kinetic energy when moving with a speed of localid="1648723450518" role="math" $2.0\times {10}^{7}m/s$.

Short answer

a. The speed of a $7.0MeV$is$3.7\times {10}^{7}m/s$.

b. The speed of a $15MeV$is$2.67\times {10}^{7}m/s$.

c. The specific type of particle is electron.

Step-by-step solution

Part (a) step 1: Given information

We need to determine the speed of a$7.0MeV$neutron.

Part (a) step 2: Simplify

Energy of motion is

$$\begin{gathered} 7.0MeV=7.0\times {10}^6\times 1.6\times {10}^{-19}Jmassm=1.6\times {10}^{-27}kg \\ v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2\times 7.0\times {10}^6\times 1.6\times {10}^{-19}}{1.6\times {10}^{-27}}}=3.7\times {10}^{7}m/s \end{gathered}$$

Part (b) step 1: Given information

We need to determine the speed of a $15MeV$helium atom.

Part (b) step 2: Simplify

Energy=$15.0MeV=15.0\times {10}^6\times 1.6\times {10}^{-19}J$,

Mass= $6.64\times {10}^{-27}kg$

Velocity $v$is

$v=\sqrt{\frac{2E_k}{m}}=\sqrt{\frac{2\times 15.0\times {10}^6\times 1.6\times {10}^{-19}}{6.64\times {10}^{-27}}}=2.67\times {10}^{7}m/s$

part (c) Given information

We need to determine the specific type of particle that has $1.14MeV$of kinetic energy when moving with a speed of localid="1648723371129" $2.0\times {10}^{7}m/s$.

Part (c) step 2: Simplify

Energy= $1.14KeV=1.14\times {10}^3\times 1.6\times {10}^{-19}J$

Velocity =$2.0\times {10}^{7}m/s$

$m=\frac{2E_k}{v^2}=\frac{2\times 1.14\times {10}^3\times 1.6\times {10}^{-19}}{{(2.0\times {10}^7)}^2}=9.1\times {10}^{-31}$

Since the mass of the particle is equal to electron, hence the unknown particle is electron.