KCET · Physics · Atomic Physics
The figure shows the energy level of certain atom. When the electron de excites from \(3 E\) to \(E\), an electromagentic wave of wavelength \(\lambda\) is emitted. What is the wavelength of the electromagnetic wave emitted when the electron de excites from \(\frac{5 E}{3}\) to \(E\) ?
- A \(3 \lambda\)
- B \(2 \lambda\)
- C \(5 \lambda\)
- D \(\frac{3 \lambda}{5}\)
Answer & Solution
Correct Answer
(A) \(3 \lambda\)
Step-by-step Solution
Detailed explanation
As electron de excites from \(3 E \rightarrow E\)
\(\begin{aligned} & \therefore & h v &=3 E-E \\ & \text { or } & \frac{h c}{\lambda} &=2 E ...(i)\end{aligned}\)
When electron de excites from \(\frac{5 E}{3} \rightarrow E\)
\(\frac{h c}{\lambda^{\prime}}=\frac{5 E}{3}-E=\frac{2 E}{3}...(ii)\)
Substituting the value of \(E\) from Eq. (i) in above equation
\(\begin{aligned}
\frac{h c}{\lambda^{\prime}} &=\frac{2}{3}\left(\frac{h c}{\lambda_{2}}\right) \\
\Rightarrow \quad \lambda^{\prime} &=3 \lambda
\end{aligned}\)
\(\begin{aligned} & \therefore & h v &=3 E-E \\ & \text { or } & \frac{h c}{\lambda} &=2 E ...(i)\end{aligned}\)
When electron de excites from \(\frac{5 E}{3} \rightarrow E\)
\(\frac{h c}{\lambda^{\prime}}=\frac{5 E}{3}-E=\frac{2 E}{3}...(ii)\)
Substituting the value of \(E\) from Eq. (i) in above equation
\(\begin{aligned}
\frac{h c}{\lambda^{\prime}} &=\frac{2}{3}\left(\frac{h c}{\lambda_{2}}\right) \\
\Rightarrow \quad \lambda^{\prime} &=3 \lambda
\end{aligned}\)
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