A photon of wavelength $\lambda = 102.6\,\text{nm}$ is incident on a ground-state hydrogen atom. The atom can absorb it and transition to which excited state? (Take $hc = 1240\,\text{eV\,nm}$.)
A$n = 2$ (first excited state)
B$n = 3$ (second excited state)
C$n = 4$ (third excited state)
DIt cannot be absorbed at all
Answer & Solution
Correct answer: B. $n = 3$ (second excited state)
1. Photon energy: $E = hc/\lambda = 1240/102.6 = 12.09\,\text{eV}$.
2. From the ground state $E_1 = -13.6\,\text{eV}$, the atom needs an energy gap of exactly $E_n - E_1$ for some integer $n \geq 2$.
3. Required final energy: $E_n = E_1 + 12.09 = -13.6 + 12.09 = -1.51\,\text{eV}$.
4. From $E_n = -13.6/n^2$: $-1.51 = -13.6/n^2 \Rightarrow n^2 = 13.6/1.51 = 9.00 \Rightarrow n = 3$.
5. So a $12.09\,\text{eV}$ photon resonantly drives the $n=1 \to n=3$ transition. This is the Ly-$\beta$ line.
6. Option A would require $E = 10.2\,\text{eV}$ (Ly-$\alpha$). Option C would require $E = 12.75\,\text{eV}$. Option D forgets that quantised levels admit specific photon energies — and $12.09$ is one of them.
_Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1 + §12.5 (energy level values and absorption), p. 10–11._
Related questions
According to de Broglie's interpretation of Bohr's quantisation, the circumference of the In the Bohr model, the velocity of an electron in the $n^\text{th}$ orbit of hydrogen is gIn the Bohr model, an electron transitions from $n = 3$ to $n = 2$ in a hydrogen atom (theThe ratio of the first excitation energy of a hydrogen atom ($E_2 - E_1$) to its ionisatioAn $\alpha$-particle of energy $7.7\,\text{MeV}$ is directed head-on at a stationary gold In the Bohr model, the radius of the $n^\text{th}$ orbit of a hydrogen-LIKE atom of atomicWhen the electron in a hydrogen atom makes a transition from $n = 2$ (first excited state)Bohr's quantisation condition for the angular momentum $L$ of an electron in the $n^\text{