Structure Of Atom Question 6
Question 6 - 29 January - Shift 1
The shortest wavelength of hydrogen atom in Lyman series is $\lambda$. The longest wavelength in Balmer series of $He^{+}$is
(1) $\frac{5}{9 \lambda}$
(2) $\frac{9 \lambda}{5}$
(3) $\frac{36 \lambda}{5}$
(4) $\frac{5 \lambda}{9}$
Show Answer
Answer: (2)
Solution:
Formula: Wavelength of Emitted Photon
For $H$ : $\frac{1}{\lambda}=R_H \times 1^{2}(\frac{1}{1^{2}}-\frac{1}{\infty^{2}})$
$\frac{1}{\lambda _{He^{+}}}=R_H \times 2^{2} \times(\frac{1}{4}-\frac{1}{9})$
From (1) & (2) $\frac{\lambda _{He^{+}}}{\lambda}=\frac{9}{5}$
$\lambda _{He^{+}}=\lambda \times \frac{9}{5}$
$\lambda _{He^{+}}=\frac{9 \lambda}{5}$
