Atoms & Nuclei

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118 of 239

Question 118 of 239

If 917A˚917Å be the lowest wavelength of Lyman series then the lowest wavelength of Balmer series will be A˚Å.

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Answer: 3668

Using Rydberg's formula,
1λ=RZ2(1n21−1n22)\dfrac{1}{\lambda }=R{Z}^{2}\left(\dfrac{1}{{{n}^{2}}_{1}}-\dfrac{1}{{{n}^{2}}_{2}}\right)
For the lowest wavelength in Lyman series, n1=1,n2=∞{n}_{1}=1,{n}_{2}=\infty.
1λ=RZ2(112−1∞2)=RZ2\dfrac{1}{\lambda }=R{Z}^{2}\left(\dfrac{1}{{1}^{2}}-\dfrac{1}{{\infty }^{2}}\right)=R{Z}^{2}
For Balmer series, n1=2,n2=∞{n}_{1}=2,{n}_{2}=\infty
1λ′=RZ2(122−1∞2)=RZ24\dfrac{1}{\lambda '}=R{Z}^{2}\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{\infty }^{2}}\right)=\dfrac{R{Z}^{2}}{4}
So,
1λ′=14λ⇒λ′=4×917A˚=3668A˚\dfrac{1}{\lambda '}=\dfrac{1}{4\lambda } \Rightarrow \lambda '=4\times 917Å=3668Å^{}

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