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

Q: Can I just subtract the wavelengths directly, like $\lambda_2 = 680 - 355$?

No, you cannot. Energy is conserved, not wavelength. Because energy is inversely proportional to wavelength ($E = \frac{hc}{\lambda}$), you must add or subtract the energies, which translates to adding or subtracting the reciprocals of the wavelengths.

Q: Why do we assume no energy is lost in the process?

In these ideal quantum mechanics problems, unless specified otherwise, we assume 100% of the absorbed energy is re-emitted as photons. The atom returns to its original ground state, meaning the net energy change of the atom is zero.

Q: What if the atom emitted three photons instead of two?

The principle remains exactly the same. The total energy absorbed would equal the sum of the three emitted energies: $E_{\text{total}} = E_1 + E_2 + E_3$. This would lead to the equation $\frac{1}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2} + \frac{1}{\lambda_3}$.

Q: Do I need to convert the wavelengths from nanometers to meters before calculating?

No, it's not necessary here. Since all terms in the equation $\frac{1}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$ have the unit of inverse length, you can keep them in nanometers. Your final answer for $\lambda_2$ will naturally come out in nanometers.

Q: Why is the calculated answer $742.77$ slightly different from the option $743$?

In many competitive exams like JEE, options are often rounded to the nearest integer for simplicity. $742.77$ rounds perfectly to $743$, making it the unambiguously correct choice.

Question

A gas absorbs photon of $355 	ext{ nm}$ and emits at two wavelengths. If one of the emission is at $680 	ext{ nm}$, the other is at

Options:

(A) $1035 	ext{ nm}$

(B) $325 	ext{ nm}$

(C) $743 	ext{ nm}$

(D) $518 	ext{ nm}$

Sigma Percentile · Accepted Answer

### The Quantum Leap: A Tale of Energy Conservation

Imagine a tiny gas atom, resting peacefully in its ground state. Suddenly, a high-energy photon with a wavelength of $355 	ext{ nm}$ strikes it. The atom absorbs this photon and makes a massive quantum leap to a highly excited state.

But what goes up must come down. Instead of plummeting back to the ground state in one giant leap, our atom decides to take the scenic route. It drops down in two distinct steps, emitting a photon at each step. We are told that one of these emitted photons has a wavelength of $680 	ext{ nm}$. Our mission is to find the wavelength of the second photon.

### The Master Equation: Conservation of Energy

The fundamental rule governing this entire process is the **Law of Conservation of Energy**. The total energy the atom absorbed must be exactly equal to the total energy it emits.

Mathematically, we can write this as:

$$E_{	ext{total}} = E_1 + E_2$$

Where $E_{	ext{total}}$ is the energy of the absorbed photon, and $E_1$ and $E_2$ are the energies of the two emitted photons.

Now, we need to translate this energy equation into the language of wavelengths. Enter **Planck's Quantum Theory**, which tells us that the energy of a photon is inversely proportional to its wavelength:

$$E = \frac{hc}{\lambda}$$

Substituting this into our conservation equation, we get:

$$\frac{hc}{\lambda} = \frac{hc}{\lambda_1} + \frac{hc}{\lambda_2}$$

### The Elegance of Simplification

Look closely at the equation. The term $hc$ (Planck's constant times the speed of light) is present in every single numerator. We can elegantly divide the entire equation by $hc$, causing it to vanish completely. This leaves us with a beautifully simple relationship:

$$\frac{1}{\lambda} = \frac{1}{\lambda_1} + \frac{1}{\lambda_2}$$

**A crucial trap warning:** Notice that we are adding the *reciprocals* of the wavelengths, not the wavelengths themselves. A common silly mistake is to assume $\lambda = \lambda_1 + \lambda_2$. This is physically incorrect because energy and wavelength are inversely related!

### Final Calculation

Now, let's substitute the values given in the problem. The absorbed wavelength $\lambda$ is $355 	ext{ nm}$, and the first emitted wavelength $\lambda_1$ is $680 	ext{ nm}$.

$$\frac{1}{355} = \frac{1}{680} + \frac{1}{\lambda_2}$$

Rearranging to solve for the unknown $\lambda_2$:

$$\frac{1}{\lambda_2} = \frac{1}{355} - \frac{1}{680}$$

To subtract these fractions, we find a common denominator:

$$\frac{1}{\lambda_2} = \frac{680 - 355}{355 	imes 680}$$

$$\frac{1}{\lambda_2} = \frac{325}{241400}$$

Finally, we invert the fraction to find $\lambda_2$:

$$\lambda_2 = \frac{241400}{325} \approx 742.77 	ext{ nm}$$

Rounding to the nearest integer, we get **$743 	ext{ nm}$**. This perfectly matches option (c). The beauty of this problem lies in how a complex quantum event can be unraveled using simple algebraic conservation laws.

Animated Solution for Chemistry - Atomic Structure: A gas absorbs photon of $355 nm$ and emits at two wavelengths. If one of the emission is at $680 nm$ , the other is at

Select Answer:

Visualized Solution

\text{Energy Conservation}

\text{Planck's Quantum Theory}

\text{Simplifying the Equation}

\text{Substituting Values}

\text{Calculation}

\text{Conclusion}

The Sigma Insight: Wave Particle Duality

The Quantum Leap

The Master Equation

The Elegance of Simplification

Final Calculation

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Animated Solution for Chemistry - Atomic Structure: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } A gas absorbs photon of 355 nm and emits at two wavelengths. If one of the emission is at 680 nm, the other is at