Calculate the longest wavelength in the hydro | vedclass

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(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$.For the Lyman

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$1.216 	imes 10^{-5} \ cm$

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(B) The Rydberg formula for the hydrogen spectrum is given by $\frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Lyman series,$n_1 = 1$.The longest wavelength corresponds to the smallest energy transition,which occurs from $n_2 = 2$ to $n_1 = 1$.Substituting the values: $\frac{1}{\lambda} = 109677 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) \ cm^{-1}$.$\frac{1}{\lambda} = 109677 \left( 1 - \frac{1}{4} ight) = 109677 	imes \frac{3}{4} \ cm^{-1}$.$\frac{1}{\lambda} = 82257.75 \ cm^{-1}$.$\lambda = \frac{1}{82257.75} \ cm \approx 1.216 	imes 10^{-5} \ cm$.

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

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