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(A) From Rydberg's equation,the wavelength $\lambda$ for an electron transition is given by:$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n

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$\frac{25}{24 R}$

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(A) From Rydberg's equation,the wavelength $\lambda$ for an electron transition is given by:$\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right)$Here,the transition is from $n_i = 5$ to $n_f = 1$.Substituting the values:$\frac{1}{\lambda} = R \left( \frac{1}{1^2} - \frac{1}{5^2} \right)$$\frac{1}{\lambda} = R \left( 1 - \frac{1}{25} \right)$$\frac{1}{\lambda} = R \left( \frac{25 - 1}{25} \right)$$\frac{1}{\lambda} = \frac{24 R}{25}$Therefore,the wavelength is $\lambda = \frac{25}{24 R}$.

$A$ hydrogen atom emits a photon corresponding to an electron transition from $n=5$ to $n=1$. If $R$ is the Rydberg constant,then the wavelength of the emitted photon is

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