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NEET 2016
Exam
NEET
Year
2016
Subject
Physics
Chapter
Atomic Physics

NEET 2016 · Physics · Atomic Physics

Given the value of Rydberg constant is 10 ^ 7 m ^ - 1 , the wave...

Given the value of Rydberg constant is 107m−1{10}^{7}{m}^{- 1}, the wave number of the last line of the Balmer series in hydrogen spectrum will be:

Options

  1. A0.025×104m−10.025 \times {10}^{4}{m}^{- 1}
  2. B0.5×107m−10.5 \times {10}^{7}{m}^{- 1}
  3. C0.25×107m−10.25 \times {10}^{7}{m}^{- 1}
  4. D2.5×107m−12.5 \times {10}^{7}{m}^{- 1}

Correct answer

C. 0.25×107m−10.25 \times {10}^{7}{m}^{- 1}

Step-by-step solution

1λ=R(1n12−1n22)\dfrac{1}{\lambda} = R( 1{n}_{1}^{2} - 1{n}_{2}^{2} )
Here, [n1=2,n2=∞][ {n}_{1} = 2, {n}_{2} = \infty ]
Wavelength =1λ=R[122]=R4=1074=0.25×107m−1= \dfrac{1}{\lambda} = R\left[\dfrac{1}{{2}^{2}}\right] = \dfrac{R}{4} = \dfrac{{10}^{7}}{4} = 0.25 \times {10}^{7}{m}^{- 1}
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