Question

Among the following transitions in hydrogen and hydrogen like ion spectrum, which one emits light of longest wavelength?

• A. n = 2 to n = 1 for H
• B. n = 4 to n =3 fro Li${}^{2+}$
• C. n = 4 to n =3 for He${}^{+}$
• D. n = 5 to n = 2 for H

Answer 1

Answer: (C)

n = 4 to n =3 for He${}^{+}$

As we know that

$\frac{1}{\lambda }=R_H\times Z^2(\frac{1}{n_1^2}-\frac{1}{n_2^2})R_H=1.0974\times {10}^7{m}^{-1}$

For $H:$ n=2 to n=1

$\frac{1}{\lambda }=R_H(\frac{1}{1}-\frac{1}{4})\frac{1}{\lambda }=\frac{3}{4}{R}_H$

$\lambda =1.33R_H$

For ${Li}^{2+}$ : n=4 to n=3

$\frac{1}{\lambda }=R_H\times 9\times (\frac{1}{9}-\frac{1}{16})\frac{1}{\lambda }=\frac{7}{16}{R}_H$

$\lambda =2.28R_H$

For ${He}^{1+}$ : n=4 to n=3

$\frac{1}{\lambda }=R_H\times 4\times (\frac{1}{9}-\frac{1}{16})\frac{1}{\lambda }=\frac{7}{36}{R}_H$

$\lambda =5.14R_H$

For $H$ : n=5 to n=2

$\frac{1}{\lambda }=R_H(\frac{1}{4}-\frac{1}{25})\frac{1}{\lambda }=\frac{21}{100}{R}_H$

$\lambda =4.76R_H$

So, longest wavelength light emitted when ${He}^{1+}$ jumps from n=4 to n=3.