The transition from the state n = 4 to n = 3 in a hydrogen like atom results in ultraviolet radiatio

Q: Why does infrared have lower energy than ultraviolet?

The energy of a photon is given by $E = hc/\lambda$. Infrared light has a longer wavelength than ultraviolet light, which means it has a lower frequency and, consequently, lower energy.

Q: Do we need to know the exact value of $Z$ to solve this?

No, the atomic number $Z$ is constant for the given hydrogen-like atom. We only need to compare the relative energies of different transitions, so $Z^2$ cancels out in the comparison.

Q: What if multiple options had lower energy?

In a single-choice question, only one option will satisfy the condition. If multiple options had lower energy, the question might specify a particular series (like Paschen or Brackett) or it would be a multiple-correct question.

Q: Are all transitions ending in $n=4$ infrared?

For a hydrogen atom, transitions ending in $n=4$ belong to the Brackett series, which lies entirely in the infrared region. However, for hydrogen-like atoms with higher $Z$, the energies are scaled by $Z^2$, so they might shift to higher energy regions. But relatively speaking, they will always be lower energy than transitions ending in lower $n$ from the next adjacent level.

Q: Can we solve this without calculating the exact fractions?

Yes! The energy gap between adjacent levels ($\Delta E_{n+1 \rightarrow n}$) decreases rapidly as $n$ increases. So, $\Delta E_{2 \rightarrow 1} > \Delta E_{3 \rightarrow 2} > \Delta E_{4 \rightarrow 3} > \Delta E_{5 \rightarrow 4}$. Since $4 \rightarrow 3$ is UV, any transition between higher adjacent levels (like $5 \rightarrow 4$) will have lower energy (IR).

Question

The transition from the state $n = 4$ to $n = 3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

Options:

(A) $2 \rightarrow 1$

(B) $3 \rightarrow 2$

(C) $4 \rightarrow 2$

(D) $5 \rightarrow 4$

Sigma Percentile · Accepted Answer

### The Quantum Leap: Decoding Atomic Transitions

Imagine an atom as a microscopic staircase, where each step represents a specific energy level. Electrons can only exist on these steps, never in between. When an electron decides to take a leap from a higher step down to a lower one, it sheds its excess energy in the form of a tiny packet of light called a photon. This is the beautiful essence of the Bohr model of the atom.

### The Energy Equation

The energy of the photon emitted during this quantum leap is precisely the difference in energy between the two levels. Mathematically, for a hydrogen-like atom, this energy difference $\Delta E$ is given by the Rydberg formula:

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

Here, $n_i$ is the initial higher energy level, $n_f$ is the final lower energy level, and $Z$ is the atomic number. Notice how the energy depends on the inverse squares of these quantum numbers. This means that jumps between lower energy levels (like $2 ightarrow 1$) involve massive energy changes compared to jumps between higher energy levels (like $5 ightarrow 4$).

### The Electromagnetic Spectrum Connection

The problem tells us that the transition from $n = 4$ to $n = 3$ produces ultraviolet (UV) radiation. We know from the electromagnetic spectrum that UV light is highly energetic. Infrared (IR) radiation, on the other hand, sits on the lower energy side of the visible spectrum.

Therefore, to find the transition that produces infrared radiation, we are hunting for a jump that releases **less energy** than our UV benchmark ($4 ightarrow 3$).

### Evaluating the Options

Let's calculate the relative energy factor $\left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$ for our benchmark and the given options:

**The Benchmark (UV):**
$4 ightarrow 3$: $\frac{1}{3^2} - \frac{1}{4^2} = \frac{1}{9} - \frac{1}{16} = \frac{7}{144} \approx 0.0486$

**The Contenders:**
(a) $2 ightarrow 1$: $\frac{1}{1^2} - \frac{1}{2^2} = \frac{3}{4} = 0.75$
(b) $3 ightarrow 2$: $\frac{1}{2^2} - \frac{1}{3^2} = \frac{5}{36} \approx 0.139$
(c) $4 ightarrow 2$: $\frac{1}{2^2} - \frac{1}{4^2} = \frac{3}{16} \approx 0.1875$

Look closely at these values. They are all significantly larger than our benchmark of $0.0486$. This means these transitions release *more* energy than the UV transition, pushing them further into the extreme UV or X-ray regions. They are definitely not infrared.

**The Winner:**
(d) $5 ightarrow 4$: $\frac{1}{4^2} - \frac{1}{5^2} = \frac{1}{16} - \frac{1}{25} = \frac{9}{400} = 0.0225$

This value ($0.0225$) is strictly less than our UV benchmark ($0.0486$). Because it releases less energy, the emitted photon will have a longer wavelength, placing it squarely in the infrared region of the spectrum.

### A Quick Trick

You don't always need to calculate the exact fractions! The energy gap between adjacent levels ($\Delta E_{n+1 ightarrow n}$) shrinks rapidly as you move further away from the nucleus. So, $\Delta E_{2 ightarrow 1} > \Delta E_{3 ightarrow 2} > \Delta E_{4 ightarrow 3} > \Delta E_{5 ightarrow 4}$. If $4 ightarrow 3$ is UV, the only adjacent transition with lower energy is $5 ightarrow 4$, making it the perfect candidate for IR radiation.

Animated Solution for Physics - Atoms and Nuclei: The transition from the state $n = 4$ to $n = 3$ in a hydrogen like atom results in ultraviolet radiation. Infrared radiation will be obtained in the transition

Select Answer:

Visualized Solution

Energy of Transition

Electromagnetic Spectrum

Energy of $4 \to 3$ Transition

Checking Options (a), (b), (c)

Checking Option (d)

Conclusion

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

The Quantum Leap

The Energy Equation

The Electromagnetic Spectrum Connection

Evaluating the Options

A Quick Trick

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