91Ó°ÊÓ

The energy carried away by a photon when the electron in A jumps down from the fourth excited state to the first excited state is,

$E_{A14}=25\text{eV}$.

Energy levels in B are as follow.

$\begin{array}{rcl}{E}_{B1}& =& 2\text{eV}\\ E_{B2}& =& 8\text{eV}\\ E_{B3}& =& 19\text{eV}\end{array}$

Energy levels in C are given below.

$\begin{array}{rcl}{E}_{C1}& =& 4\text{eV}\\ E_{C2}& =& 16\text{eV}\end{array}$

Energy levels in D are as follow.

$\begin{array}{rcl}{E}_{D1}& =& 3\text{eV}\\ E_{D2}& =& 12\text{eV}\\ E_{D3}& =& 27\text{eV}\end{array}$

Energy levels in E are,

$\begin{array}{rcl}{E}_{E1}& =& 5\text{eV}\\ E_{E2}& =& 20\text{eV}\end{array}$

A photon can be absorbed by an electron in a particular energy state if the energy carried by the photon is exactly equal to the difference in energy of a higher energy state to the one the electron is in. The electron will then excite to the higher level.

By inspection it can be checked that the energy of the emitted photon $E_{A14}$ is exactly equal to the difference in energy between the third excited state and the ground state of D.

$\begin{array}{rcl}{E}_{D3}-E_{D1}& =& 27\text{eV -}3\text{eV}\\ & =& 24\text{eV}\\ & =& E_{A14}\end{array}$

Hence, the electron in the ground state (n = 1) of D absorbed the photon and jumped to n = 3 of D.