KCET · Physics · Atomic Physics
Three energy levels of hydrogen atom and the corresponding wavelength of the emitted radiation due to different electron transition are as shown. Then,
- A \(\lambda_3=\frac{\lambda_1 \lambda_2}{\lambda_1+\lambda_2}\)
- B \(\lambda_1=\frac{\lambda_2 \lambda_3}{\lambda_2+\lambda_3}\)
- C \(\lambda_2=\lambda_1+\lambda_3\)
- D \(\lambda_2=\frac{\lambda_1 \lambda_3}{\lambda_1+\lambda_3}\)
Answer & Solution
Correct Answer
(D) \(\lambda_2=\frac{\lambda_1 \lambda_3}{\lambda_1+\lambda_3}\)
Step-by-step Solution
Detailed explanation
From the given diagram,
\(\begin{gathered}E_2-E_1=\frac{h c}{\lambda_1} \\ E_3-E_2=\frac{h c}{\lambda_3} \\ E_3-E_1=\frac{h c}{\lambda_2}\end{gathered}\)
Adding Eq. (i) and Eq. (ii), we get
\(\begin{aligned} E_3-E_1 & =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_3} \\ \Rightarrow \quad \frac{h c}{\lambda_2} & =\frac{h c}{\lambda_1}+\frac{h c}{\lambda_3}\end{aligned}\)
[From Eq. (iii)]
\(\begin{aligned} & \Rightarrow \quad \frac{1}{\lambda_2}=\frac{1}{\lambda_1}+\frac{1}{\lambda_3} \\ & \Rightarrow \quad \lambda_2=\frac{\lambda_1 \lambda_3}{\lambda_1+\lambda_3}\end{aligned}\)
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