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If the first line in the Lyman series has wavelength $\lambda$, then the first line in Balmer series has the wavelength
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Verified Answer
The correct answer is:
$\frac{27}{5} \lambda$
Wavelength for Lyman series is calculated as
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
For first line of Lyman series, $n=2$
$\therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \frac{1}{\lambda}=\frac{3 R}{4}$
Wavelength of Balmer series is calculated as
$\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
For first line of Balmer series, $n=3$
$\begin{array}{ll}\therefore & \frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{36} R \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} \times \frac{4}{3 \lambda} \quad \text { [from Eq. (i)] } \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{27 \lambda} \Rightarrow \lambda^{\prime}=\frac{27 \lambda}{5}\end{array}$
$\frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{n^2}\right)$
For first line of Lyman series, $n=2$
$\therefore \quad \frac{1}{\lambda}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right) \Rightarrow \frac{1}{\lambda}=\frac{3 R}{4}$
Wavelength of Balmer series is calculated as
$\frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{n^2}\right)$
For first line of Balmer series, $n=3$
$\begin{array}{ll}\therefore & \frac{1}{\lambda^{\prime}}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right) \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{36} R \Rightarrow \frac{1}{\lambda^{\prime}}=\frac{5}{36} \times \frac{4}{3 \lambda} \quad \text { [from Eq. (i)] } \\ \Rightarrow & \frac{1}{\lambda^{\prime}}=\frac{5}{27 \lambda} \Rightarrow \lambda^{\prime}=\frac{27 \lambda}{5}\end{array}$
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