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(A) The Lyman series corresponds to transitions ending at $n_1 = 1$. The $1^{st}$ Lyman line is $n_2 = 2 \rightarrow n_1 = 1$. The $3^{rd}$ Lyman line

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$\frac{16}{3R}$

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(A) The Lyman series corresponds to transitions ending at $n_1 = 1$. The $1^{st}$ Lyman line is $n_2 = 2 \rightarrow n_1 = 1$. The $3^{rd}$ Lyman line is $n_2 = 4 \rightarrow n_1 = 1$. Given the transition is from the $3^{rd}$ Lyman line $(n_2 = 4)$ to the $1^{st}$ Lyman line $(n_2 = 2)$: Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} \right]$ For hydrogen,$Z = 1$,$n_1 = 2$,and $n_2 = 4$: $\frac{1}{\lambda} = R \left[ \frac{1}{2^2} - \frac{1}{4^2} \right]$ $\frac{1}{\lambda} = R \left[ \frac{1}{4} - \frac{1}{16} \right]$ $\frac{1}{\lambda} = R \left[ \frac{4-1}{16} \right] = R \left[ \frac{3}{16} \right]$ $\lambda = \frac{16}{3R}$

If in a hydrogen atom,an electron jumps from the $3^{rd}$ Lyman line to the $1^{st}$ Lyman line,then the obtained wavelength is:

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