An electron in a hydrogen atom transitions from $n = 3$ to $n = 2$. Calculate the wavelength of the emitted radiation. (Rydberg constant $R = 1.097 \times 10^7$ m$^{-1}$.)
A$486$ nm
B$410$ nm
C$1216$ nm
D$656$ nm
Answer & Solution
Correct answer: D. $656$ nm
Rydberg formula: $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right)$.
$\dfrac{1}{\lambda} = 1.097 \times 10^7 \times \left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 1.097 \times 10^7 \times \dfrac{5}{36} \approx 1.524 \times 10^6$ m$^{-1}$.
$\lambda \approx 6.56 \times 10^{-7}$ m $= 656$ nm.
This is the **H-α line** of the Balmer series — the deep-red line in the visible hydrogen spectrum. Option D ($1216$ nm) is Lyman-α (UV, $n=2 \to 1$); options A and B are higher Balmer lines (H-δ and H-β).
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 atomicWhen the electron in a hydrogen atom makes a transition from $n = 2$ (first excited state)