Question

Consider hypothetical hydrogen like atom. The wavelength in A for the spectral lines for translation from n = p to n = 1 are given by $\lambda =\frac{1500p^2}{p^2-1}$, where p = 2,3,4 ...... (given hc = 12400 eV/A)

• A. The wavelength of the least energetic and the most energetic photons in this senrs is 2000 A,1500 A
• B. Difference between energies of fourth and third orbit is 0.40 eV
• C. Energy of second orbit is 6.2 eV
• D. The ionisation potential of this element is 8.27 V

Answer 1

The correct option is B Difference between energies of fourth and third orbit is 0.40 eV

$\begin{array}{c}\text{Given,}\\ \text{a hypothetical hydrogen like atom}\\ \text{wavelength of spectal lime(}\lambda )=\frac{1500p^2}{p^2-1}\\ \text{where}p=2,3,4\end{array}$

$\text{and}hc=12400\frac{ev}{\dot{A}}$

$\begin{array}{c}\text{(a) For least energetic photon}\\ \begin{array}{cc}n=1& \text{to}p=2\\ \lambda & =\frac{1500\times (2{)}^2}{4-1}\\ \lambda & =2000\dot{A}\end{array}\end{array}$

$\begin{array}{c}\text{For most energetic photon}\\ n=1\text{to}p=\infty \\ \begin{array}{cc}\lambda =\frac{1500\times p^2}{p^2-1}& =\frac{1500}{(1-\frac{1}{p^2})}\\ & =\frac{1500}{(1-\frac{1}{\infty })}=1500\dot{A}\end{array}\end{array}$

${\lambda }_{\infty -1}=1500\dot{A}$

$E_{\infty }-E_1=\frac{12400}{1500}ev$

$\begin{array}{c}\text{and we knows Energy at.}\left(\infty \right)=0\\ E_{\infty }=0\\ \begin{array}{c}{E}_1=-\frac{12400}{1500}ev\\ E_1=-8.267ev\end{array}\end{array}$

$\begin{array}{cc}\text{for}{E}_2,& \\ {\lambda }_2-{\lambda }_1& =2000\dot{A}\\ E_2-E_1& =\frac{12400}{2000}\left(ev\right)\\ E_2-E_1& =6.2\\ E_2& =6.2-8.2\\ E_2& =-2.0\end{array}$

$\begin{array}{c}\text{for}{E}_3\\ {\lambda }_{3-1}=\frac{1500}{\left(1\frac{1}{9}\right)}=1687.54\dot{A}\\ E_3-E_1=\frac{12400}{1687.5}ev\\ E_3=-0.95ev\\ \text{Similary}{E}_4=0.55\end{array}$

$\begin{array}{c}\text{hence our option (A) is correct}\\ \text{option (D) is also correct}\\ \begin{array}{cc}\text{and}{E}_4-E_3& =+0.95-0.55\\ & =0.40ev\end{array}\end{array}$