When the electron in a hydrogen atom makes a transition from $n = 2$ (first excited state) DIRECTLY back to $n = 1$ (ground state), the energy of the EMITTED photon is
A$3.40\,\text{eV}$
B$10.2\,\text{eV}$
C$12.09\,\text{eV}$
D$13.6\,\text{eV}$
Answer & Solution
Correct answer: B. $10.2\,\text{eV}$
1. Bohr's third postulate: when an electron drops from a higher energy $E_i$ to a lower energy $E_f$, a photon of energy $h\nu = E_i - E_f$ is emitted.
2. Use $E_n = -13.6/n^2\,\text{eV}$.
3. Initial: $E_2 = -13.6/4 = -3.40\,\text{eV}$.
4. Final: $E_1 = -13.6/1 = -13.6\,\text{eV}$.
5. Photon energy: $h\nu = E_2 - E_1 = -3.40 - (-13.6) = 10.20\,\text{eV}$.
6. This is the first Lyman line (often denoted $\text{Ly-}\alpha$). Option A is just $|E_2|$. Option C is $E_3 - E_1 = 12.09\,\text{eV}$ (Ly-β). Option D is the ionisation energy itself.
_Source: NCERT Class 12 Physics Part 2, Ch 12, §12.4.1 (Eq. 12.10 + numerical values), p. 10._
Related questions
A photon of wavelength $\lambda = 102.6\,\text{nm}$ is incident on a ground-state hydrogenAccording 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 atomicBohr's quantisation condition for the angular momentum $L$ of an electron in the $n^\text{