Question

Which is the smallest wavelength that will be observed in spectra of $H{e}^{+}$ ion?

• A. $24.4nm$
• B. $28.8nm$
• C. $22.2nm$
• D. $30.6nm$

Answer 1

The correct option is A $24.4nm$

Energy levels of ${He}^{+}$ are $-54.4eV,-13.6eV,-6.04eV,-3.4eV$ ..............
Therefore, ${He}^{+}$ was initially in $n=2$ state, then excited to $n=4$.
Now, the smallest wavelength that will be observed will correspond to large energy difference.
$\Rightarrow$ De-Excitation from $n=4$ to $n-1$ and the corresponding wavelength is
$\frac{1}{\lambda }=(1.097\times {10}^7)\left(4\right)(1-\frac{1}{16})=4.1\times {10}^7\Rightarrow \lambda =24.3\times {10}^{-9}m$