ExamBro
ExamBro
AP EAMCET · PHYSICS · Atomic Physics

If \(\lambda_1\) and \(\lambda_2\) are the wavelength of the photons emitted, when electrons in the \(n^{\text {th }}\) orbit of hydrogen atom fall to first excited state and ground state respectively, then the value of n is

  1. A \(\sqrt{\frac{2\left(\lambda_2-\lambda_1\right)}{2 \lambda_2-\lambda_1}}\)
  2. B \(\sqrt{\frac{2 \lambda_2-\lambda_1}{2\left(\lambda_2-\lambda_1\right)}}\)
  3. C \(\sqrt{\frac{4 \lambda_2-\lambda_1}{4\left(\lambda_2-\lambda_1\right)}}\)
  4. D \(\sqrt{\frac{4\left(\lambda_2-\lambda_1\right)}{\left(4 \lambda_2-\lambda_1\right)}}\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(\sqrt{\frac{4\left(\lambda_2-\lambda_1\right)}{\left(4 \lambda_2-\lambda_1\right)}}\)

Step-by-step Solution

Detailed explanation

We have, \(\lambda_1=\frac{h c}{\Delta E_{n \rightarrow 2}} \text { and } \lambda_2=\frac{h c}{\Delta E_{n \rightarrow 1}}\)…
Same subject
Explore more questions on app