Atoms & Nuclei

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139 of 239

Question 139 of 239

Taking the wavelength of first Balmer line in hydrogen spectrum (n=3(n=3 to n=2)n=2) as 660nm660nm , the wavelength of the 2nd{2}^{nd} Balmer line (n=4(n=4 to n=2)n=2) will be :

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Answer: (B) 488.9nm488.9nm

From Rydberg’s equation,
1λ=R(1n2−1n22)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{n}^{2}}-\dfrac{1}{{n}_{2}^{2}}\right)
1660×10−9=R(122−132)\dfrac{1}{660\times {10}^{-9}}=R\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{3}^{2}}\right)
1λ=R(122−142)\dfrac{1}{\lambda }=R\left(\dfrac{1}{{2}^{2}}-\dfrac{1}{{4}^{2}}\right)
λ600×10−9=59×4×16×412\dfrac{\lambda }{600\times {10}^{-9}}=\dfrac{5}{9\times 4}\times \dfrac{16\times 4}{12}
λ=488.9nm\lambda =488.9 nm

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