Answer
See the explanation
Work Step by Step
Let's start with the Rydberg formula:
\[\Delta E = -13.6 Z^2\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)\]
Here, ∆E is the energy change, \(n_f\) is the final energy level, and \(n_i\) is the initial energy level. In an atom, an electron can move from one level to another by absorbing or emitting energy, in this case, a photon.
For the hydrogen atom (\(H\)) transition from \(n_i=5\) to \(n_f=2\), the energy change (∆E) can be calculated as follows:
\[\Delta E_{\text{H}} = -13.6\cdot 1^2\cdot \left( \frac{1}{2^2} - \frac{1}{5^2} \right)\]
\[\Delta E_{\text{H}} = -13.6 \left( \frac{1}{4} - \frac{1}{25} \right)\]
\[\Delta E_{\text{H}} = -2.856\]
Now, we have to find out the final energy level (\(n_f\)) for the helium ion (He⁺) if the initial energy level (\(n_i\)) is 4. The energy of the photon emitted should be the same as that of the hydrogen atom transition, leading to the same blue light. So, we equate the energy change for the He⁺ ion to that of the hydrogen atom:
\[-13.6\cdot 2^2\left( \frac{1}{n_f^2} - \frac{1}{4^2} \right) = -2.856\]
To find \(n_f\), we isolate \(n_f\) on one side of the equation:
\[\frac{1}{n_f^2} - \frac{1}{4^2} = \frac{2.856}{13.6\cdot 4}\]
\[n_f^2 = \frac{1}{\frac{1}{4} - \frac{1}{25} + \frac{1}{4^2}}\]
\[n_f = \sqrt{\frac{1}{0.115}}\approx 3\]
Round \(n_f\) to the nearest whole number, since energy levels are quantized in whole numbers for electrons in atoms.
The solution of the above equation will yield the desired value of \(n_f\) for the He⁺ ion that would result in the emission of the same blue light as in the hydrogen atom transition.
