Test Index

CBSE Class 12 Physics 2023 Outside Delhi Set 1 Paper

© examsnet.com
Question : 11 of 35
Marks: +1, -0
A hydrogen atom makes a transition from n=5n=5 to n=1n=1 orbit. The wavelength of photon emitted is λ\lambda. The wavelength of photon emitted when it makes a transition from n=5n=5 to n=2n=2 orbit is
Solution:     👈: Video Solution
In 1  st  1^{\;\text{st}\;} case
    1λ=R[112−152]\;\;\frac{1}{\lambda}= R\left[\frac{1}{1^2}-\frac{1}{5^2}\right]
   or,       1λ=R[1−125]=R[2425]\;\text{ or, }\;\;\;\frac{1}{\lambda}= R\left[1-\frac{1}{25}\right] = R\left[\frac{24}{25}\right]
∴  R=2524×1λ\therefore \; R = \frac{25}{24} \times \frac{1}{\lambda}
In 2  nd  2^{\;\text{nd}\;} case
1λ′=R[122−152]\frac{1}{\lambda'} = R\left[\frac{1}{2^2}-\frac{1}{5^2}\right]
   or,       1λ′=R[14−125]=R×21100\;\text{ or, }\;\;\;\frac{1}{\lambda'} = R\left[\frac{1}{4}-\frac{1}{25}\right] = R \times \frac{21}{100}
   or,       1λ′=2524×1λ×21100\;\text{ or, }\;\;\;\frac{1}{\lambda'} = \frac{25}{24} \times \frac{1}{\lambda} \times \frac{21}{100}
   or,       1λ′=2196λ\;\text{ or, }\;\;\;\frac{1}{\lambda'} = \frac{21}{96\lambda}
∴  λ′=9621λ=327λ\therefore \; \lambda' = \frac{96}{21}\lambda = \frac{32}{7}\lambda
© examsnet.com
Go to Question:
Prev Question
Next Question