Modern Physics 1 Question 36
33. The wavelength of light emitted in the visible region by $He^{+}$ions after collisions with $H$ atoms is
(a) $6.5 \times 10^{-7} m$
(b) $5.6 \times 10^{-7} m$
(c) $4.8 \times 10^{-7} m$
(d) $4.0 \times 10^{-7} m$
(2008, 4M)
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Solution:
- Visible light lies in the range, $\lambda _1=4000 \AA$ to $\lambda _2=7000 \AA$. Energy of photons corresponding to these wavelengths (in $eV$ ) would be :
$$ \begin{aligned} & E _1=\frac{12375}{4000}=3.09 eV \\ & E _2=\frac{12375}{7000}=1.77 eV \end{aligned} $$
From energy level diagram of $He^{+}$atom, we can see that in transition from $n=4$ to $n=3$, energy of photon released will lie between $E _1$ and $E _2$.
$$ \begin{aligned} \Delta E _{43} & =-3.4-(-6.04) \\ & =2.64 eV \end{aligned} $$
Wavelength of photon corresponding to this energy,
$$ \begin{aligned} \lambda & =\frac{12375}{2.64} \AA=4687.5 \AA \\ & =4.68 \times 10^{-7} m \end{aligned} $$
