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(B) The first excited state of a hydrogen atom corresponds to $n_1 = 2$.Given that the electron jumps from orbit $n_2 = n$ to $n_1 = 2$,we use the Ryd

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$\sqrt{\frac{4 \lambda R}{\lambda R-4}}$

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(B) The first excited state of a hydrogen atom corresponds to $n_1 = 2$.Given that the electron jumps from orbit $n_2 = n$ to $n_1 = 2$,we use the Rydberg formula:$\frac{1}{\lambda} = R \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$Substituting the values:$\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{n^2} \right] = R \left[ \frac{1}{4} - \frac{1}{n^2} \right]$Rearranging the terms to solve for $n$:$\frac{1}{\lambda} = \frac{R}{4} - \frac{R}{n^2}$$\frac{R}{n^2} = \frac{R}{4} - \frac{1}{\lambda} = \frac{R \lambda - 4}{4 \lambda}$Taking the reciprocal:$\frac{n^2}{R} = \frac{4 \lambda}{R \lambda - 4}$$n^2 = \frac{4 \lambda R}{R \lambda - 4}$$n = \sqrt{\frac{4 \lambda R}{R \lambda - 4}}$

In a hydrogen atom,if an electron in the orbit with principal quantum number $n$ jumps to the first excited state,the wavelength of the emitted photon is $\lambda$. Then the value of $n$ is (where $R$ is the Rydberg constant).

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