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CBSE Class 12 Physics 2019 Delhi Set 3 Paper

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Question : 3 of 8
Marks: +1, -0
SECTION - B
Obtain the expression for the ratio of the de Broglie wavelengths associated with the electron orbiting in the second and third excited states of hydrogen atom.
Solution:  
2πr=nλ2 \pi r = n \lambda
For second excited state (n=3)(n=3)
r=0.529(n)2 A˚r = 0.529 (n)^2 \, \text{Å}
=0.529(3)2= 0.529 (3)^2
⇒2π(0.529)(3)2=3λ2\Rightarrow 2 \pi (0.529) (3)^2 = 3 \lambda_2
For third excited state n=4n=4
r=0.529(4)2r = 0.529 (4)^2
⇒2π(0.529)(4)2=4λ3\Rightarrow 2 \pi (0.529) (4)^2 = 4 \lambda_3
⇒  3λ24λ3=  (3)2(4)2\Rightarrow \; \frac{3 \lambda_2}{4 \lambda_3} = \; \frac{(3)^2}{(4)^2}
  λ2λ3=3:4\; \frac{\lambda_2}{\lambda_3} = 3:4
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