A hydrogen atom transitions from an excited s | vedclass

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \r

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$\sqrt{\frac{\lambda R}{\lambda R-1}}$

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(A) The Rydberg formula for the wavelength of emitted radiation is given by $\frac{1}{\lambda} = R \left( \frac{1}{n_{f}^{2}} - \frac{1}{n_{i}^{2}} \right)$.Here,the final state is the ground state,so $n_{f} = 1$,and the initial excited state is $n_{i} = n$.Substituting these values,we get $\frac{1}{\lambda} = R \left( \frac{1}{1^{2}} - \frac{1}{n^{2}} \right) = R \left( 1 - \frac{1}{n^{2}} \right)$.Rearranging the equation: $\frac{1}{\lambda R} = 1 - \frac{1}{n^{2}}$.This implies $\frac{1}{n^{2}} = 1 - \frac{1}{\lambda R} = \frac{\lambda R - 1}{\lambda R}$.Taking the reciprocal and square root,we find $n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.

$A$ hydrogen atom transitions from an excited state to the ground state by emitting a photon of wavelength $\lambda$. If $R$ is the Rydberg constant,the principal quantum number $n$ of the excited state is:

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