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  • The first member of the Balmer series of the hydrogen atom has a wavelength of 6561 . The wavelength of the second member of the Balmer series (in nm ) is _______ .

The first member of the Balmer series of the hydrogen atom has a wavelength of 6561 \AA . The wavelength of the second member of the Balmer series (in nm ) is _______ .
Option: 1 486
Option: 2 972
Option: 3 986
Option: 4 586
 

Answers (1)

best_answer

\begin{aligned} \frac{1}{\lambda} &=\mathrm{R} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) \\ \frac{1}{\lambda_{1}} &=R(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{5 \mathrm{R}}{36} \\ \frac{1}{\lambda_{2}} &=\mathrm{R}(1)^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right)=\frac{3 \mathrm{R}}{16} \\ \frac{\lambda_{2}}{\lambda_{1}} &=\frac{20}{27} \\ \lambda_{2} &=\frac{20}{27} \times 6561 \mathrm{A}=4860 \mathrm{A} \\ &=486 \mathrm{nm} \end{aligned}

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