Question

An atom makes a transition on receiving energy from an external source, as shown in the figure. Depict the lines of possible spectrum and calculate the wavelength of any two spectral lines. (Energy levels: \(n=1\) at 0, \(n=2\) at 4.86 eV, \(n=3\) at 6.67 eV, \(n=4\) at 8.80 eV.)

Hint:

An atom excited to n=4 gives n(n-1)/2 = 6 lines. Photon energy is the level difference; find each wavelength from lambda = hc/E.

Answer 1

Step 1: Number of possible spectral lines.
The atom is excited up to \(n=4\). While returning it can jump between any pair of levels. Number of lines \(=\dfrac{n(n-1)}{2}=\dfrac{4\times3}{2}=6\).
Step 2: List the six emission transitions and their energies.
\(4\to1:\ 8.80-0=8.80\) eV
\(4\to2:\ 8.80-4.86=3.94\) eV
\(4\to3:\ 8.80-6.67=2.13\) eV
\(3\to1:\ 6.67-0=6.67\) eV
\(3\to2:\ 6.67-4.86=1.81\) eV
\(2\to1:\ 4.86-0=4.86\) eV
These six downward arrows are the possible spectral lines.
Step 3: Formula for wavelength.
Energy of a photon \(E=\dfrac{hc}{\lambda}\Rightarrow \lambda=\dfrac{hc}{E}\), with \(h=6.6\times10^{-34}\) J s, \(c=3\times10^8\) m/s, \(1\,\text{eV}=1.6\times10^{-19}\) J.
\(hc=6.6\times10^{-34}\times3\times10^8=1.98\times10^{-25}\) J m.
Step 4: Wavelength of the \(4\to1\) line (E = 8.80 eV).
\(E=8.80\times1.6\times10^{-19}=1.408\times10^{-18}\) J.
\[\lambda_{4\to1}=\frac{1.98\times10^{-25}}{1.408\times10^{-18}}=1.406\times10^{-7}\,\text{m}=1406\ \text{\AA}\]Step 5: Wavelength of the \(2\to1\) line (E = 4.86 eV).
\(E=4.86\times1.6\times10^{-19}=7.776\times10^{-19}\) J.
\[\lambda_{2\to1}=\frac{1.98\times10^{-25}}{7.776\times10^{-19}}=2.546\times10^{-7}\,\text{m}=2546\ \text{\AA}\]\[\boxed{6\ \text{spectral lines};\ \lambda_{4\to1}\approx1406\,\text{\AA},\ \lambda_{2\to1}\approx2546\,\text{\AA}}\]