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: Count the lines.
From the top level \(n=4\) the electron can cascade down; the total number of distinct emission lines equals the number of level pairs, \(\binom{4}{2}=6\).
Step 2: Use a compact energy-conversion constant.
Combine the data as \(hc=1240\) eV\(\cdot\)nm (i.e. \(\lambda(\text{nm})=1240/E(\text{eV})\)). This is just \(hc\) expressed in eV\(\cdot\)nm and speeds up the arithmetic.
Step 3: Take two different lines than the boxed pair? Same two, cross-checked.
For the largest jump \(4\to1\), \(E=8.80\) eV: \(\lambda=\dfrac{1240}{8.80}=140.9\) nm \(\approx 1409\) \(\text{\AA}\).
For \(3\to1\), \(E=6.67\) eV: \(\lambda=\dfrac{1240}{6.67}=185.9\) nm \(\approx 1859\) \(\text{\AA}\).
Step 4: Interpret.
A bigger energy gap gives a shorter wavelength, so \(4\to1\) is the most energetic (shortest \(\lambda\)) line and lies in the ultraviolet region. The small \(1\) percent difference from the boxed value comes only from rounding \(hc\).
\[\boxed{6\ \text{lines};\ \lambda_{4\to1}\approx1409\,\text{\AA},\ \lambda_{3\to1}\approx1859\,\text{\AA}}\]