Chapter 27, Problem 56P

To determine

Can the photon can be absorbed by hydrogen atom present in the Sun and what energy states will the hydrogen atom transition occurs.

Answer to Problem 56P

Solution:

• Yes, photon can be absorbed by hydrogen atom present in the Sun.

• The hydrogen atom will jump from $n=1\text{and}n=3$.

Explanation of Solution

Given:

The helium atom makes a transition from the $n=6\text{and}n=2$ state.

Formula Used:

$\begin{array}{l}{E}_n=-\frac{Z^2(13.6\text{eV})}{n^2}\\ E=E_6-E_2=-E_n\left[\left(\frac{1}{6^2}-\frac{1}{2^2}\right)\right]\end{array}$

Calculation:

When the helium atom is singly ionized, it will appear like having an atom with 1 electron and 2 proton (Z = 2), now the value of $E_n$ can be calculated as by using the formula $E_n=-\frac{Z^2(13.6\text{eV})}{n^2}$

So, $E_n=-\frac{Z^2(13.6\text{eV})}{n^2}=-\frac{2^2\left(13.6\text{eV}\right)}{n^2}=\frac{-54.4\text{eV}}{n^2}$

Energy of the photon in the transition $n=6\text{and}n=2$ can be calculated as

$\begin{array}{l}E=E_6-E_2=-54.4\text{eV}\left[\left(\frac{1}{6^2}-\frac{1}{2^2}\right)\right]\\ E=E_6-E_2=-(54.4\text{eV)}\left[\left(\frac{1}{6^2}-\frac{1}{2^2}\right)\right]=12.1\text{eV}\end{array}$

The energy $12.1\text{eV}$ is the energy difference for $n=1\text{and}n=3$ in hydrogen so the photon can be absorbed by a hydrogen atom.

The energy states between $n=1\text{and}n=3$, the hydrogen atom transition occur.