Chapter 24.1, Problem 4SSC

Interpretation Introduction

Interpretation:

The approximate energy of alpha, beta and gamma radiation needs to be converted to Joule.

Concept introduction:

The are many units to measure energy, such as, joule, $\text{MeV}$, Calorie etc. A joule is defined as the energy used when $\text{1}\text{kg}$ is moved to $\text{1}\text{m}$ using the force of 1 Newton.

Answer to Problem 4SSC

The energies are $8\times {10}^{-13}\text{J}$ (alpha radiation), $0.08\times {10}^{-13}\text{J}$ (beta radiation) and $\text{1}\text{.6}\times {\text{10}}^{-13}\text{J}$ (gamma radiation)

Explanation of Solution

The approximate energy of alpha radiation, beta radiation and gamma radiation is $5\text{MeV}$, $0.05\text{to}\text{1MeV}$ and $\text{1MeV}$ respectively.

The conversion factor is $1\text{MeV}\text{=}\text{1}\text{.6}\times {\text{10}}^{-13}\text{J}$.

For alpha radiation,

$\begin{array}{c}\text{Energy}\text{in}\text{Joule}\text{=}5\cancel{\text{MeV}}\times \frac{1.6\times {10}^{-13}\text{J}}{1\cancel{\text{MeV}}}\\ =8\times {10}^{-13}\text{J}\end{array}$

For beta radiation,

$\begin{array}{c}\text{Energy}\text{in}\text{Joule}\text{=}0.05\cancel{\text{MeV}}\times \frac{1.6\times {10}^{-13}\text{J}}{1\cancel{\text{MeV}}}\\ =0.08\times {10}^{-13}\text{J}\end{array}$

For gamma radiation,

$\begin{array}{c}\text{Energy}\text{in}\text{Joule}\text{=}1\cancel{\text{MeV}}\times \frac{1.6\times {10}^{-13}\text{J}}{1\cancel{\text{MeV}}}\\ =1.6\times {10}^{-13}\text{J}\end{array}$

Conclusion

The conversion factor to convert $\text{MeV}$ to J is $1\text{MeV}\text{=}\text{1}\text{.6}\times {\text{10}}^{-13}\text{J}$.