A free hydrogen atom after absorbing a photon of wavelength λ_a gets excited from the state n = 1 to

Q: Why do we use $13.6 \text{ eV}$ in the energy equation?

This is the ground state ionization energy of the Hydrogen atom. The energy of any level $n$ in Hydrogen is given by $-13.6/n^2 \text{ eV}$. The difference between two levels naturally brings out this constant.

Q: Why did we divide the two equations instead of calculating $\lambda_a$ first?

Efficiency! While you *could* calculate $\lambda_a$ and then find $\lambda_e$, dividing the equations eliminates the constants $hc$ and $13.6$, saving you from tedious calculations and reducing the chance of silly mistakes.

Q: How do we know the change in momentum of the atom is equal to $h/\lambda$?

This comes from the conservation of momentum. Initially, the atom is at rest. When it absorbs or emits a photon (which carries momentum $p = h/\lambda$), the atom must recoil with an equal and opposite momentum to keep the total momentum conserved.

Q: Why is the kinetic energy proportional to $1/n^2$?

In Bohr's model, the velocity of an electron is proportional to $1/n$. Since Kinetic Energy is $\frac{1}{2}mv^2$, squaring the velocity makes the kinetic energy proportional to $1/n^2$.

Q: Is the recoil energy of the atom significant here?

Great question! For a hydrogen atom, its mass is so large compared to the energy of the photon that the recoil kinetic energy is negligibly small. Therefore, we can safely assume that almost 100% of the photon's energy goes into the electronic transition.

Question

A free hydrogen atom after absorbing a photon of wavelength $\lambda_a$ gets excited from the state $n = 1$ to the state $n = 4$. Immediately after that the electron jumps to $n = m$ state by emitting a photon of wavelength $\lambda_e$. Let the change in momentum of atom due to the absorption and the emission are $\Delta p_a$ and $\Delta p_e$, respectively. If $\lambda_a/\lambda_e = \frac{1}{5}$. Which of the option(s) is/are correct?
[Use $hc = 1242 	ext{ eV nm}$; $1 	ext{ nm} = 10^{-9} 	ext{ m}$, $h$ and $c$ are Planck's constant and speed of light, respectively]

Options:

(A) $\lambda_e = 418 	ext{ nm}$

(B) The ratio of kinetic energy of the electron in the state $n = m$ to the state $n = 1$ is $\frac{1}{4}$

(C) $m = 2$

(D) $\Delta p_a/\Delta p_e = \frac{1}{2}$

Sigma Percentile · Accepted Answer

### The Quantum Setup: Absorption and Emission

Imagine a hydrogen atom resting peacefully in its ground state, where the principal quantum number is $n = 1$. Suddenly, it absorbs a photon of wavelength $\lambda_a$. This influx of energy excites the electron, propelling it all the way up to the $n = 4$ state.

But the electron doesn't stay there for long. Almost immediately, it drops back down to an intermediate state $n = m$. To shed the excess energy, it emits a new photon, this time with a wavelength of $\lambda_e$. Our mission is to decode this quantum dance, find the mystery state $m$, and verify the physical properties of this transition.

### The Master Equation: Rydberg's Formula

To relate the wavelengths of the photons to the energy levels of the hydrogen atom, we rely on the famous Rydberg formula. The energy difference $\Delta E$ in any transition is given by:

$$ \Delta E = \frac{hc}{\lambda} = 13.6 \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) 	ext{ eV} $$

Let's set up the equations for both processes. For the absorption process, the electron jumps from $n=1$ to $n=4$:

$$ \frac{hc}{\lambda_a} = 13.6 \left( \frac{1}{1^2} - \frac{1}{4^2} ight) $$

For the emission process, the electron drops from $n=4$ to $n=m$:

$$ \frac{hc}{\lambda_e} = 13.6 \left( \frac{1}{m^2} - \frac{1}{4^2} ight) $$

### The Algebraic Elegance: Finding the Mystery State

We are given a crucial piece of information: the ratio of the wavelengths $\lambda_a / \lambda_e = \frac{1}{5}$. Instead of calculating the wavelengths individually, we can divide our two Rydberg equations. This brilliant move eliminates the constants $hc$ and $13.6$, leaving us with a clean algebraic expression:

$$ \frac{\lambda_a}{\lambda_e} = \frac{\frac{1}{m^2} - \frac{1}{16}}{1 - \frac{1}{16}} = \frac{1}{5} $$

Let's solve this step-by-step. The denominator simplifies to $\frac{15}{16}$. Multiplying this across to the right side gives:

$$ \frac{1}{m^2} - \frac{1}{16} = \frac{15}{16} 	imes \frac{1}{5} = \frac{3}{16} $$

Now, we simply add $\frac{1}{16}$ to both sides:

$$ \frac{1}{m^2} = \frac{3}{16} + \frac{1}{16} = \frac{4}{16} = \frac{1}{4} $$

Taking the square root reveals our mystery state: **$m = 2$**. This confirms that option (C) is correct!

### Verifying the Wavelength

Now that we know $m=2$, let's check option (A) by calculating the exact emission wavelength $\lambda_e$. Substituting $m=2$ back into our emission equation:

$$ \frac{hc}{\lambda_e} = 13.6 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 13.6 	imes \frac{3}{16} 	ext{ eV} $$

Using the given value $hc = 1242 	ext{ eV nm}$, we can solve for $\lambda_e$:

$$ \lambda_e = \frac{1242 	imes 16}{13.6 	imes 3} \approx 487 	ext{ nm} $$

Since option (A) claims $\lambda_e = 418 	ext{ nm}$, it is incorrect.

### Kinetic Energy and Momentum: The Final Checks

Let's evaluate option (B), which discusses the kinetic energy. In the Bohr model, the kinetic energy of an electron is inversely proportional to the square of the principal quantum number ($KE \propto \frac{1}{n^2}$). Therefore, the ratio of the kinetic energy in state $m=2$ to the ground state $n=1$ is:

$$ \frac{KE_2}{KE_1} = \frac{1^2}{2^2} = \frac{1}{4} $$

This perfectly matches option (B)!

Finally, let's look at the change in momentum for option (D). By conservation of momentum, the recoil momentum of the atom equals the momentum of the photon, which is $p = \frac{h}{\lambda}$. The ratio of the momentum changes is simply the inverse ratio of the wavelengths:

$$ \frac{\Delta p_a}{\Delta p_e} = \frac{\frac{h}{\lambda_a}}{\frac{h}{\lambda_e}} = \frac{\lambda_e}{\lambda_a} = 5 $$

Option (D) suggests the ratio is $\frac{1}{2}$, which is incorrect.

By systematically applying the Rydberg formula and Bohr's postulates, we have successfully navigated this quantum puzzle. The correct options are indeed **(B) and (C)**.

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Visualized Solution

$Energy Levels of Hydrogen Atom$

$Rydberg Equation for Energy Transitions$

$Equations for Absorption and Emission$

$Utilizing the Given Wavelength Ratio$

$Calculating the Final State m$

$Calculating Emission Wavelength λ_{e}$

$Ratio of Kinetic Energies$

$Ratio of Change in Momentum$

$Conclusion$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Quantum Setup

The Master Equation

The Algebraic Elegance

Verifying the Wavelength

Kinetic Energy and Momentum

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Comprehension Passage

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy Levels of Hydrogen Atom

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Rydberg Equation for Energy Transitions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equations for Absorption and Emission

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Utilizing the Given Wavelength Ratio

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating the Final State m

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Emission Wavelength λe​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Kinetic Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Change in Momentum

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion