quantum numbers

Question

Approximate wavelength (nm) of the photon emitted when an electron in hydrogen atom drops from n=4 to n=2 (Balmer series):

RankGuru · Accepted Answer

1. **Rydberg Equation**: $\frac{1}{\lambda} = R_H Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.
2. **Constants**: $R_H \approx 1.097 	imes 10^7$ m⁻¹. For H, $Z=1$.
3. **Transition**: $n_1=2$ (Balmer series) and $n_2=4$.
4. **Calculation**:
   - $\frac{1}{\lambda} = 1.097 	imes 10^7 \left( 1/4 - 1/16 ight) = 1.097 	imes 10^7 \left( 3/16 ight)$
   - $\lambda \approx 486.2 	imes 10^{-9}$ m.
5. **Unit conversion**: **$486$ nm**.
6. **Identification**: This is the $H_\beta$ line. The $3 	o 2$ transition ($H_\alpha$) occurs at $656$ nm.

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