Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r = lambda_1

Q: Why do we subtract the final energy from the initial energy?

The energy of the emitted photon must be a positive value. Since the electron goes from a higher (less negative) energy state to a lower (more negative) energy state, calculating $E_{\text{initial}} - E_{\text{final}}$ gives the exact positive energy difference carried away by the photon.

Q: Does the negative sign in the energy levels mean the energy is actually negative?

Yes, bound states in atoms or molecules have negative total energy. This simply means that energy must be supplied to the system to free the electron. The 'zero' of energy is typically set at an infinite distance from the nucleus.

Q: Why is $\lambda$ inversely proportional to $\Delta E$?

According to the Planck-Einstein relation, the energy of a photon is $E = h\nu$. Since wave speed $c = \nu\lambda$, we can substitute frequency to write $E = \frac{hc}{\lambda}$. Here, $h$ and $c$ are constants, so $E$ is strictly proportional to $\frac{1}{\lambda}$.

Q: What if the electron absorbed a photon instead of emitting one?

If it absorbed a photon, it would transition from a lower energy level to a higher one. The energy of the absorbed photon would still be the absolute difference between the two levels, but the final energy state would be greater than the initial energy state.

Q: Why didn't we need the actual numerical values of $h$ and $c$?

Because we were asked for a ratio of wavelengths. When you divide $\lambda_1$ by $\lambda_2$, the constants $h$ and $c$ cancel out perfectly in the numerator and denominator, leaving only the inverse ratio of the energy differences.

Question

Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r = \frac{\lambda_1}{\lambda_2}$ is given by

Options:

(A) $r = \frac{2}{3}$

(B) $r = \frac{3}{4}$

(C) $r = \frac{1}{3}$

(D) $r = \frac{4}{3}$

Sigma Percentile · Accepted Answer

### The Energy Level Landscape

Imagine an electron residing within a molecule, navigating a ladder of distinct energy states. In our given problem, we are presented with a visual map of these states: four specific energy levels denoted as $-E$, $-\frac{4}{3}E$, $-2E$, and $-3E$.

The negative signs simply indicate that the electron is in a bound state; it would require positive energy to completely free it from the molecule. We are tasked with analyzing two specific downward jumps, or transitions, made by the electron. When an electron drops from a higher energy level to a lower one, it sheds its excess energy by emitting a photon of light. The first transition emits a photon with wavelength $\lambda_1$, and the second emits a photon with wavelength $\lambda_2$.

### The Core Principle: Energy and Wavelength

To solve this, we need the fundamental bridge connecting the macroscopic world of wavelengths to the quantum world of energy levels. This bridge is the Planck-Einstein relation combined with the wave equation:

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

Here, $\Delta E$ is the energy difference between the two levels, $h$ is Planck's constant, and $c$ is the speed of light. Because $h$ and $c$ are constants, we can clearly see a beautiful inverse relationship: the wavelength $\lambda$ is inversely proportional to the energy gap $\Delta E$.

$$ \lambda \propto \frac{1}{\Delta E} $$

This means a massive energy drop produces a highly energetic photon with a very short wavelength, while a tiny energy drop produces a low-energy photon with a long wavelength.

### Calculating the Energy Gaps

Let's meticulously calculate the energy gap for each transition. The energy of the emitted photon is always the initial (higher) energy minus the final (lower) energy.

**For the first transition ($\lambda_1$):**
The electron leaps from the top level ($-E$) down to the third level ($-2E$).

$$ \Delta E_1 = E_{	ext{initial}} - E_{	ext{final}} $$
$$ \Delta E_1 = -E - (-2E) $$
$$ \Delta E_1 = -E + 2E = E $$

So, the photon associated with $\lambda_1$ carries away an energy of exactly $E$.

**For the second transition ($\lambda_2$):**
Now, the electron takes a smaller step, jumping from the top level ($-E$) to the second level ($-\frac{4}{3}E$).

$$ \Delta E_2 = E_{	ext{initial}} - E_{	ext{final}} $$
$$ \Delta E_2 = -E - \left(-\frac{4}{3}Eight) $$
$$ \Delta E_2 = -E + \frac{4}{3}E = \frac{1}{3}E $$

The photon associated with $\lambda_2$ carries away an energy of $\frac{1}{3}E$.

### The Final Ratio

We are asked to find the ratio $r = \frac{\lambda_1}{\lambda_2}$. Armed with our inverse proportionality rule, we know that the ratio of the wavelengths is simply the inverse ratio of their corresponding energy gaps:

$$ \frac{\lambda_1}{\lambda_2} = \frac{\Delta E_2}{\Delta E_1} $$

Now, we just substitute the energy gaps we calculated:

$$ r = \frac{\frac{1}{3}E}{E} $$

The $E$ terms elegantly cancel out, leaving us with our final answer:

$$ r = \frac{1}{3} $$

This tells us that because the first transition involved three times as much energy as the second transition, its emitted photon has exactly one-third the wavelength.

Animated Solution for Physics - Atoms: Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths $r = \frac{λ _{1}}{λ _{2}}$ is given by

Select Answer:

Visualized Solution

Visual Anchor

Logic Bridge

Raw Setup for $λ_{1}$

Atomic Compute for $λ_{1}$

Raw Setup for $λ_{2}$

Atomic Compute for $λ_{2}$

Final Answer

The Way Forward

The Sigma Insight: Bohr Model and Hydrogen Spectra

The Energy Level Landscape

The Core Principle

Calculating the Energy Gaps

The Final Ratio

Similar Questions

The atomic hydrogen emits a line spectrum consisting of various series. Which series of hydrogen atomic spectra is lying in the visible region ?

Animated Solution for Physics - Atoms: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Some energy levels of a molecule are shown in the figure. The ratio of the wavelengths r=λ2​λ1​​ is given by