Atomic Structure - Result Question 98
####74. The wavelength corresponding to maximum energy for hydrogen is $91.2 nm$. Find the corresponding wavelength for $He^{+}$ion.
(2003, 2M)
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Answer:
Correct Answer: 74. $\left(2.725 \times 10^{6} M^{-1}\right)$
Solution:
- The general Rydberg’s equation is
$$ \begin{aligned} & \bar{v}=\frac{1}{\lambda}=R(Z)^{2}\left(\frac{1}{n _1^{2}}-\frac{1}{n _2^{2}}\right) \ \Rightarrow & \frac{1}{\lambda} \propto Z^{2} \ \Rightarrow \quad & \frac{\lambda\left(He^{+}\right)}{\lambda(H)}=\frac{Z(H)^{2}}{Z\left(He^{+}\right)^{2}}=\frac{1}{4} \ \Rightarrow \quad & \lambda\left(He^{+}\right)=\frac{\lambda(H)}{4}=\frac{91.2}{4} nm=22.8 nm \end{aligned} $$
