A gas absorbs a photon of 355 nm and emits at two wavelengths | Chemca Solution

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Question 43

A gas absorbs a photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at:

(a) 1035 nm

(b) 325 nm

(c) 743 nm

(d) 518 nm

Detailed Step-by-Step Solution

This problem is based on the Law of Conservation of Energy. According to this principle, the total energy of the absorbed photon must be equal to the sum of the energies of the emitted photons.

Step 1: Set up the energy equation

Let \( E_{\text{abs}} \) be the energy absorbed, and \( E_{1} \) and \( E_{2} \) be the energies of the two emitted photons. We can write:

\( E_{\text{abs}} = E_{1} + E_{2} \)

We know from Planck's quantum theory that energy \( E = \frac{hc}{\lambda} \). Substituting this into our equation gives:

\( \frac{hc}{\lambda_{\text{abs}}} = \frac{hc}{\lambda_{1}} + \frac{hc}{\lambda_{2}} \)

Step 2: Simplify and plug in the values

We can cancel out Planck's constant (\(h\)) and the speed of light (\(c\)) from both sides of the equation. This simplifies to a relationship between the wavelengths:

\( \frac{1}{\lambda_{\text{abs}}} = \frac{1}{\lambda_{1}} + \frac{1}{\lambda_{2}} \)

Given the values:

• Absorbed wavelength (\( \lambda_{\text{abs}} \)) = \( 355 \text{ nm} \)
• First emitted wavelength (\( \lambda_{1} \)) = \( 680 \text{ nm} \)
• Second emitted wavelength (\( \lambda_{2} \)) = ?

\( \frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_{2}} \)

Step 3: Solve for the unknown wavelength (\(\lambda_2\))

Rearrange the equation to isolate \( \frac{1}{\lambda_2} \):

\( \frac{1}{\lambda_{2}} = \frac{1}{355} - \frac{1}{680} \)

Find a common denominator or use cross-multiplication:

\( \frac{1}{\lambda_{2}} = \frac{680 - 355}{355 \times 680} \)

\( \frac{1}{\lambda_{2}} = \frac{325}{241400} \)

Now, invert the fraction to solve for \( \lambda_2 \):

\( \lambda_{2} = \frac{241400}{325} \)

\( \lambda_{2} \approx 742.76 \text{ nm} \)

Rounding off to the nearest whole number gives us \( 743 \text{ nm} \).

Conclusion: The wavelength of the other emitted photon is approximately 743 nm. Therefore, the correct option is (c) 743 nm.

Mastering Atomic Structure Concepts

Questions involving photon absorption and multi-step emission rely heavily on Planck's Quantum Theory and the principles of energy conservation. If you find numericals like these tricky, we recommend brushing up on your foundational concepts. You can read our comprehensive, easy-to-understand notes on the Structure of Atom Class 11 Chemistry.

Chemca is dedicated to providing students with top-tier resources. Make sure to explore our full syllabus guides:

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