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Question 52 of 57

Q.Calculate the wavelength of the first two lines in Balmer series of hydrogen atom.

Maharashtra MsbshseMaharashtra HSC (MSBSHSE) Board 2024Subjective· 3mImportance★★★★★
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First two Balmer lines from the Rydberg formula with n1=2n_1=2.

The Balmer series wavelengths satisfy

1λ=R(122−1n2),n=3,4,5,…,R=1.097×107 m−1\frac{1}{\lambda} = R\left(\frac{1}{2^2}-\frac{1}{n^2}\right), \qquad n=3,4,5,\dots, \qquad R=1.097\times10^7\ \text{m}^{-1}

First line (HαH_\alpha, n=3n=3):

1λ=R(14−19)=R⋅536=5×1.097×10736≈1.524×106 m−1\frac{1}{\lambda} = R\left(\frac{1}{4}-\frac{1}{9}\right) = R\cdot\frac{5}{36} = \frac{5\times1.097\times10^7}{36} \approx 1.524\times10^6\ \text{m}^{-1}

λ≈6.564×10−7 m=656.4 nm\lambda \approx 6.564\times10^{-7}\ \text{m} = 656.4\ \text{nm}

Second line (HβH_\beta, n=4n=4): …

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