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Exercises · 8.3

Q.(a) A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.

(b) Show that the total number of lines in the emission spectrum is n(n−1)2\dfrac{n(n-1)}{2} and compute the total number of possible lines in the emission spectrum.
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Step 1 (part a). The absorbed photon's energy is E=hcλ=(6.63×10−34)(3×108)97.5×10−9≈2.04×10−18 J≈12.75E=\dfrac{hc}{\lambda}=\dfrac{(6.63\times10^{-34})(3\times10^8)}{97.5\times10^{-9}}\approx2.04\times10^{-18}\ \text{J}\approx12.75 eV.

Step 2. Since the atom starts in the ground state (E1=−13.6E_1=-13.6 eV), the final state has energy En=E1+12.75=−13.6+12.75≈−0.85E_n=E_1+12.75=-13.6+12.75\approx-0.85 eV. Comparing with En=−13.6/n2E_n=-13.6/n^2 eV: −13.6/n2≈−0.85⇒n2≈16⇒n=4-13.6/n^2\approx-0.85\Rightarrow n^2\approx16\Rightarrow n=4.

Step 3 (part b). From an excited state nn, an electron can make a downward transition to any of the n−1n-1 lower levels directly below it, but every distinct pair of levels among the nn occupied/reachable levels corresponds to one possible spectral line - i.e. the number of ways to choose 2 levels out of nn is (n2)=n(n−1)2\binom{n}{2}=\dfrac{n(n-1)}{2}.

Step 4. With n=4n=4: total lines =4×32=6=\dfrac{4\times3}{2}=6.

Step 5. These 6 lines correspond to the transitions 4→34\to3, 4→24\to2, 4→14\to1, 3→23\to2, 3→13\to1, 2→12\to1 - exactly the complete set of pairwise transitions among levels 1 through 4.

✓Final answer

(a) n=4n=4.

(b) Total possible emission lines =n(n−1)/2=4×3/2=6=n(n-1)/2=4\times3/2=6.

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