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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{Rch}{n^2}$.The energy of the emitted photon is the diff

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

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(B) The energy of an electron in the $n$-th orbit of a hydrogen atom is given by $E_n = -\frac{Rch}{n^2}$.The energy of the emitted photon is the difference between the energy of the excited state $(E_n)$ and the ground state $(E_1)$:$E_{	ext{photon}} = E_n - E_1 = \frac{hc}{\lambda}$Substituting the energy values:$-\frac{Rch}{n^2} - (-\frac{Rch}{1^2}) = \frac{hc}{\lambda}$Dividing both sides by $hc$:$-\frac{R}{n^2} + R = \frac{1}{\lambda}$Rearranging the terms:$R - \frac{1}{\lambda} = \frac{R}{n^2}$$\frac{\lambda R - 1}{\lambda} = \frac{R}{n^2}$$n^2 = \frac{\lambda R}{\lambda R - 1}$Therefore,$n = \sqrt{\frac{\lambda R}{\lambda R - 1}}$.

$A$ hydrogen atom in an excited state transitions to the ground state by emitting a photon of wavelength $\lambda$. The value of the principal quantum number $n$ of the excited state is:
($R$: Rydberg constant)

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$A$ hydrogen atom in an excited state transitions to the ground state by emitting a photon of wavelength $\lambda$. The value of the principal quantum number $n$ of the excited state is:($R$: Rydberg constant)