Question

An ionic atom equivalent to hydrogen atom has wavelength equal to 1/4 ${}^{th}$ of the wave lengths of hydrogen lines, the ion is:

• A. ${He}^{+}$
• B. ${Li}^{+}$
• C. ${Ne}^{+8}$
• D. ${Na}^{+}$

Answer 1

Answer: (A)

${He}^{+}$

We know $v^{-}=1/\lambda =R_H[1/n_1^2-1/n_2^2]z^2$

0r $\lambda =\frac{1}{R_H[1/n_1^2-1/n_2^2]z^2}$

For hydrogen atom value of $z$ is 1 and for Helium atom value of z is 2

Thus ${\lambda }_H=\frac{1}{R_H[1/n_1^2-1/n_2^2]}$

and ${\lambda }_{H{e}^{+}}=\frac{1}{R_H[1/n_1^2-1/n_2^2]2^2}=\frac{1}{R_H[1/n_1^2-1/n_2^2]4}=\frac{{\lambda }_H}{4}$