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The Rydberg formula for the wavelength of radiation emitted when an electron in a hydrogen atom jumps from level $n_2$ to level $n_1$ (with $n_2 > n_1$) is:

A$\lambda = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$
B$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$
C$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_2^2} - \dfrac{1}{n_1^2}\right)$
D$\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1} - \dfrac{1}{n_2}\right)$
Answer & Solution
Correct answer: B. $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$
The standard Rydberg form for hydrogen is $\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)$ with $n_2 > n_1$, where $R = 1.097 \times 10^7\,\text{m}^{-1}$. Since $n_2 > n_1$ this expression is positive, giving a real wavelength. The form in option A would yield a negative right-hand side and is therefore wrong.
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