Chapter 30, Problem 22P

To determine

The wavelength of gamma photon emitted when the nucleus decays from the excited state to ground state.

Answer to Problem 22P

Solution:

The wavelength of gamma photon emitted is $2.585\times {10}^{-12}\text{m}$.

Explanation of Solution

Given Info:

The ${}_3{}^7\text{L}\text{i}$ has an excited state 0.48 MeV above the ground state.

Formula used:

The photon energy is $E=h\nu$, where $h=6.62608\times {10}^{-34}\text{Js}$ is Plank’s constant and $\nu -$ frequency of emitted photon.

Calculation:

We have

$\begin{array}{l}E=h\nu \\ E=h\left(\frac{c}{\lambda }\right)\\ \lambda =\frac{hc}{E}\\ \lambda =\frac{\left(6.62608\times {10}^{-34}\text{Js}\right)\left(3\times {10}^8\text{m/s}\right)}{0.48\text{MeV}}\\ \lambda =\frac{\left(6.62608\times {10}^{-34}\text{Js}\right)\left(3\times {10}^8\text{m/s}\right)}{0.48\times 1.602\times {10}^{-19}\times {10}^6\text{J}}\\ \lambda =2.585\times {10}^{-12}\text{m}\end{array}$