If our eyes could see a slightly wider region of the electromagnetic spectrum, we would see a fifth line in the Balmer series em

Question

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ission spectrum. Calculate the wavelength λ associated with the fifth line

1 answer:

castortr0y [3K] · Answer 1 · 2026-04-02T14:18:14+03:00

Response: The wavelength tied to the fifth line is 397 nm

Clarification:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of emitted radiation

E= energy

For the fifth line in the hydrogen atom's spectrum within the Balmer series, the transition occurs from n=2 to n=7.

Applying Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation

Rydberg's Constant = $R_H$ $10973731.6m^{-1}$

$n_f$ = Higher quantum level = 7

$n_i$ = Lower quantum level = 2 (within Balmer series)

Substituting the values into the equation above, we obtain

$\frac{1}{\lambda}=10973731.6\times \left(\frac{1}{2^2}-\frac{1}{7^2} \right )\times 1^2$

$\frac{1}{\lambda}=2.52\times 10^{6}m$

$\lambda}=3.97\times 10^{-7}m=397 nm$ $1nm=10^{-9}m$

Therefore, the wavelength λ corresponding to the fifth line is 397 nm