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Calculate the wavelength (in nm) of the spectral line of Lyman series for $\mathrm{n}=2$. (Rydberg constant $=1.1 \times 10^7 \mathrm{~m}^{-1}$ )
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The correct answer is:
$121$
For Lymen $n_1=1$ and $n_2=2$
And wavelength associated with electron jumping from any state to the first state according to Lyman series and it can be expressed as
$\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{2^2}\right)$
i.e. $\lambda=\frac{1}{R \times\left(\frac{4-1}{4}\right)}=\frac{1}{1.1 \times 10^7 \times \frac{3}{4}}=1.215 \times 10^{-7}=121 \mathrm{~nm}$
And wavelength associated with electron jumping from any state to the first state according to Lyman series and it can be expressed as
$\frac{1}{\lambda}=R\left(\frac{1}{1}-\frac{1}{2^2}\right)$
i.e. $\lambda=\frac{1}{R \times\left(\frac{4-1}{4}\right)}=\frac{1}{1.1 \times 10^7 \times \frac{3}{4}}=1.215 \times 10^{-7}=121 \mathrm{~nm}$
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