Question 42

The smallest wavelength of Lyman series is 91 nm. The difference between the largest wavelengths of Paschen and Balmer series is nearly __ nm.

Smallest Lyman wavelength = 91 nm → $R = 1/(91 \times 10^{-9})$.

Largest Balmer: $3 \to 2$: $\frac{1}{\lambda_B} = R(1/4-1/9) = 5R/36$. $\lambda_B = 36/(5R) = 36 \times 91/5 = 655.2$ nm.

Largest Paschen: $4 \to 3$: $\frac{1}{\lambda_P} = R(1/9-1/16) = 7R/144$. $\lambda_P = 144/(7R) = 144 \times 91/7 = 1872$ nm.

Difference = $1872 - 655.2 = 1216.8 \approx 1217$ nm.

The answer is Option 4: 1217 nm.

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