Calculate the shortest wavelength in the hydr | vedclass

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Question

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)

Vedclass · Accepted Answer

$911.7 	imes 10^{-8} 	ext{ cm}$

Vedclass · Answer

(A) The Rydberg formula for the wavelength of emitted radiation is given by: $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series, the transition occurs to the ground state, so $n_1 = 1$.For the shortest wavelength, the transition must occur from the highest possible energy level, i.e.,$n_2 = \infty$.Substituting these values into the formula: $\frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{\infty^2} ight) = R_H (1 - 0) = R_H$.Therefore, $\lambda = \frac{1}{R_H} = \frac{1}{109677 	ext{ cm}^{-1}}$.Calculating the value: $\lambda \approx 9.117 	imes 10^{-6} 	ext{ cm} = 911.7 	imes 10^{-8} 	ext{ cm}$.

Calculate the shortest wavelength in the hydrogen spectrum emission of the Lyman series $(R_H = 109677 \text{ cm}^{-1})$.

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