The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2

Q: Why do we use the Rydberg formula instead of $E = hc/\lambda$?

You can absolutely use $E = hc/\lambda$! First, find the energy difference $\Delta E = 13.6 (1/1^2 - 1/2^2) = 10.2 \text{ eV}$. Then use $\lambda = 1242 / \Delta E$ to get the wavelength directly in nanometers. The Rydberg formula is just a direct mathematical combination of these two steps.

Q: Why is my calculated answer slightly different from the options?

This is a classic JEE scenario! The exact value depends on the constants used. If you use $hc = 1242 \text{ eV}\cdot\text{nm}$, you get $121.76 \text{ nm}$ (closer to $121.8$). If you use $hc = 1240 \text{ eV}\cdot\text{nm}$, you get $121.5 \text{ nm}$. Always choose the closest option.

Q: What does the negative sign mean if I do $n_i^2 - n_f^2$?

The Rydberg formula is typically written as $1/n_f^2 - 1/n_i^2$ for emission (where $n_i > n_f$) to ensure the wavelength is positive. If you get a negative value, it simply implies energy was absorbed (an upward transition), but wavelength itself is a physical distance and must be positive.

Q: Do I need to memorize the value of the Rydberg constant?

Yes, it is highly recommended. Memorize $R \approx 1.097 \times 10^7 \text{ m}^{-1}$. Even better, memorize its reciprocal: $1/R \approx 911 \text{ \AA} = 91.1 \text{ nm}$. This will save you immense calculation time!

Q: Which spectral series does this transition belong to?

Any transition that ends at the ground state ($n=1$) belongs to the Lyman series. Since this is from $n=2$ to $n=1$, it is the first line (or alpha line) of the Lyman series, which falls in the ultraviolet region.

Question

The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n=2$ to $n=1$ state is

Options:

(A) 121.8 nm

(B) 194.8 nm

(C) 490.7 nm

(D) 913.3 nm

Sigma Percentile · Accepted Answer

The quantum world is a place of sudden, discrete changes. Unlike a ball rolling down a smooth hill, an electron in an atom doesn't gradually lose energy. It jumps. It teleports from one allowed energy state to another, and in that fleeting moment of transition, it communicates with the universe by releasing a packet of pure light—a photon.

In this problem, we are witnessing one of the most famous quantum leaps in physics: an electron in a hydrogen atom dropping from the first excited state ($n=2$) down to the ground state ($n=1$). Our mission is to decode the message it sends by calculating the exact wavelength of the emitted photon.

### Visualizing the Quantum Leap

Imagine the hydrogen atom as a microscopic staircase. The bottom step is the ground state, $n=1$. This is where the electron is most stable, tightly bound to the nucleus. The next step up is the first excited state, $n=2$.

Our electron is currently standing on the $n=2$ step. It's energetic, but it wants to return home to the ground state. When it finally makes that downward leap, the law of conservation of energy dictates that the energy it loses cannot simply vanish. Instead, it is converted into a photon that shoots off at the speed of light. The energy of this photon is exactly equal to the energy difference between the two steps.

### The Master Equation: The Rydberg Formula

To find the wavelength of this photon, we bring out a heavy hitter from our physics toolkit: the **Rydberg Formula**.

$$ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight) $$

This elegant equation is a mathematical bridge. On the right side, we have the pure quantum numbers representing the architecture of the atom ($n_i$ for the initial state and $n_f$ for the final state). On the left side, we have the physical property of the light we can measure in a laboratory: its wavelength ($\lambda$). The constant $R$ is the Rydberg constant, a fundamental number in atomic physics, approximately equal to $1.097 	imes 10^7 	ext{ m}^{-1}$.

### Executing the Calculation

Let's carefully substitute our specific values into the formula. The electron starts at $n_i = 2$ and ends at $n_f = 1$.

$$ \frac{1}{\lambda} = 1.097 	imes 10^7 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) $$

Now, we simplify the quantum bracket. This is the heart of the energy difference.

$$ \frac{1}{\lambda} = 1.097 	imes 10^7 \left( 1 - \frac{1}{4} ight) $$
$$ \frac{1}{\lambda} = 1.097 	imes 10^7 	imes \frac{3}{4} $$

We now have an expression for $\frac{1}{\lambda}$. To find the wavelength itself, we must take the reciprocal of the entire right side. This is a common place where students make a silly mistake, so proceed with caution!

$$ \lambda = \frac{4}{3 	imes 1.097 	imes 10^7} $$

Calculating the denominator gives us $3.291 	imes 10^7$. Dividing $4$ by this number yields:

$$ \lambda \approx 1.215 	imes 10^{-7} 	ext{ m} $$

In atomic physics, meters are far too large a unit. We typically express wavelengths of visible and ultraviolet light in nanometers ($	ext{nm}$). To convert, we multiply by $10^9$.

$$ \lambda = 121.5 	ext{ nm} $$

Looking at our options, the closest value is **$121.8 	ext{ nm}$**. The slight difference arises from the exact precision of the constants used (like using $hc = 1242 	ext{ eV}\cdot	ext{nm}$ instead of the Rydberg constant directly), but in a multiple-choice scenario, $121.8 	ext{ nm}$ is the undeniable winner.

### The Pro-Tip: The Energy Method

While the Rydberg formula is powerful, JEE toppers often use a faster, more intuitive method based on energy differences.

We know the energy of an electron in the $n^{	ext{th}}$ state of hydrogen is $E_n = -\frac{13.6}{n^2} 	ext{ eV}$.
The energy difference between $n=2$ and $n=1$ is:
$$ \Delta E = 13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} ight) = 13.6 	imes \frac{3}{4} = 10.2 	ext{ eV} $$

Now, we use the legendary shortcut formula connecting energy in electron-volts to wavelength in nanometers:
$$ \lambda = \frac{1242}{\Delta E} $$
$$ \lambda = \frac{1242}{10.2} \approx 121.76 	ext{ nm} $$

This method is not only faster but also lands us exactly on the option provided in the exam!

### The Way Forward

This specific transition—from $n=2$ to $n=1$—is not just a random calculation. It is the very first line of the **Lyman Series**. Any electron transition in a hydrogen atom that terminates at the ground state ($n=1$) belongs to this series. Because the energy gap down to the ground state is so massive, all Lyman series photons carry high energy and fall squarely in the invisible **Ultraviolet (UV)** region of the electromagnetic spectrum.

Every time you solve a problem like this, you aren't just doing algebra; you are decoding the fundamental light signatures of the universe!

Animated Solution for Physics - Atoms and Nuclei: The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from $n = 2$ to $n = 1$ state is

Select Answer:

Visualized Solution

$n = 2 \to n = 1$ Transition

Photon Emission

Rydberg Formula

Substituting Values

Simplifying the Bracket

Isolating $λ$

Final Calculation

The Lyman Series

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

Visualizing the Quantum Leap

The Master Equation

Executing the Calculation

The Pro-Tip

The Way Forward

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Animated Solution for Physics - Atoms and Nuclei: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The wavelength of the photon emitted by a hydrogen atom when an electron makes a transition from n=2 to n=1 state is