Question

A hypothetical atom has only three energy levels. The ground level has energy, $E_1=-8eV$. The two excited states have energies, $E_2=-6eV$ and $E_3=-2eV$. Then which of the following wavelength will not be present in the emission spectrum of this atom?

• A. $620nm$
• B. $207nm$
• C. $465nm$
• D. $310nm$

Answer 1

The correct option is C $465nm$

$\begin{array}{cc}\text{Given}-& \ast \text{Energy level}1:E_1=-8eV\\ & \ast E_2=-6eV,\ast E_3=-2eV\end{array}$

$\text{we have only three Spectral lines possible -}$

$\text{we know that}[\Delta E=\frac{hc}{\lambda }],where$ $\begin{array}{cc}\Delta E& =\text{change in energy}\\ \lambda & =\text{wavelength}\end{array}$

* For ${\lambda }_1$

$\begin{array}{cc}\Delta E=-2& -(-8)eV=6ev=\frac{hc}{{\lambda }_1}\\ \Rightarrow {\lambda }_1=207nm& \end{array}$

* For ${\lambda }_2$

$\begin{array}{cc}\Delta E=& -2-(-6)eV=4eV=\frac{hc}{{\lambda }_2}\\ \Rightarrow {\lambda }_2=310nm& \end{array}$

* For ${\lambda }_3$

$\begin{array}{cc}\Delta E& =-6-(-8)ev=2ev=\frac{hc}{{\lambda }_3}\\ & \Rightarrow {\lambda }_3=620nm\end{array}$

$\text{Hence, we don't have any spectral line of}\lambda =465nm$

$\text{Hence, option(c) is correct.}$.