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BITSAT 2009
Exam
BITSAT
Year
2009
Subject
Physics
Chapter
Atomic Physics

BITSAT 2009 · Physics · Atomic Physics

Taking Rydberg's constant R_ H =1.097 × 10^ 7 m , first and second...

Taking Rydberg's constant RH=1.097×107m\mathrm{R}_{\mathrm{H}}=1.097 \times 10^{7} \mathrm{ m}, first and second wavelength of Balmer series in hydrogen spectrum is

Options

  1. A2000A,3000A2000 \mathrm{ A}, 3000 \mathrm{ A}
  2. B1575A,2960A1575 \mathrm{ A}, 2960 \mathrm{ A}
  3. C6529A,4280A6529 \mathrm{ A}, 4280 \mathrm{ A}
  4. D6552A,4863A6552 \mathrm{ A}, 4863 \mathrm{ A}

Correct answer

D. 6552A,4863A6552 \mathrm{ A}, 4863 \mathrm{ A}

Step-by-step solution

1λ=R[1n12−1n22].Forfirstwavelength,n1=2,n2=3\frac{1}{\lambda}=\mathrm{R} \left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]. For first wavelength, \mathrm{n}_{1} =2, \mathrm{n}_{2}=3\Rightarrow \lambda_{1}=6563 \AA.Forsecondwavelength,n1=2,n2=4⇒λ2=4861A˚. For second wavelength, \mathrm{n}_{1}=2, \mathrm{n}_{2}=4 \Rightarrow \lambda_{2}=4861 \AA
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