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Physics Question on Atomic Spectra

The longest wavelength associated with Paschen series is: (Given RH=1.097×107R_H = 1.097 \times 10^7 SI unit)

A

1.094×10−61.094 \times 10^{-6} m

B

2.973×10−62.973 \times 10^{-6} m

C

3.646×10−63.646 \times 10^{-6} m

D

1.876×10−61.876 \times 10^{-6} m

Answer

1.876×10−61.876 \times 10^{-6} m

Explanation

Solution

For the longest wavelength in the Paschen series:
1λ=R[1n12−1n22]\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]
For the longest wavelength, n1=3n_1 = 3 and n2=4n_2 = 4.
1λ=R[132−142]\frac{1}{\lambda} = R \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]
1λ=R[19−116]\frac{1}{\lambda} = R \left[ \frac{1}{9} - \frac{1}{16} \right]
1λ=R⋅16−9144=R⋅7144\frac{1}{\lambda} = R \cdot \frac{16 - 9}{144} = R \cdot \frac{7}{144}
Now, substitute R=1.097×107R = 1.097 \times 10^7:
1λ=7×1.097×107144\frac{1}{\lambda} = \frac{7 \times 1.097 \times 10^7}{144}
λ=1447×1.097×107=1.876×10−6 m\lambda = \frac{144}{7 \times 1.097 \times 10^7} = 1.876 \times 10^{-6} \, \text{m}