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

Q: Why does the energy gap decrease as we go to higher orbits?

The energy of an orbit is inversely proportional to the square of the principal quantum number ($E_n \propto -\frac{1}{n^2}$). As $n$ increases, the difference between consecutive squares decreases, making the energy levels bunch closer together.

Q: Do I need to know the exact value of Rydberg's constant to solve this?

Not at all! Since we are only comparing the relative energies of different transitions in the same atom, the constant factors like $13.6 \text{ eV}$ or $R_H$ cancel out. We only need to compare the $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)$ terms.

Q: What if the question asked for X-rays instead of Infrared?

X-rays have much higher energy than UV. In that case, you would look for a transition with a proportionality factor significantly larger than the reference UV transition, such as the $2 \rightarrow 1$ transition.

Q: Is it necessary to calculate the exact decimal values for the fractions?

While exact decimals help, you can often compare fractions by finding a common denominator or just estimating. For example, $\frac{1}{16} - \frac{1}{25}$ is clearly much smaller than $\frac{1}{9} - \frac{1}{16}$ just by looking at the denominators.

Q: Does this logic apply to atoms other than Hydrogen?

Yes, the question specifies a 'hydrogen-like atom' (like $\text{He}^+$, $\text{Li}^{2+}$). The energy formula includes a $Z^2$ term, but since we are comparing transitions within the same atom, $Z^2$ remains constant and doesn't affect the comparison.

Question

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

Options:

(A) $2 \rightarrow 1$

(B) $3 \rightarrow 2$

(C) $4 \rightarrow 2$

(D) $5 \rightarrow 4$

Sigma Percentile · Accepted Answer

### The Electromagnetic Puzzle

Imagine you are an electron orbiting a nucleus. You are not allowed to just float anywhere; you must stay in specific, quantized energy levels. When you jump from a higher energy level to a lower one, you must shed the excess energy. You do this by emitting a photon of light.

The question presents us with a fascinating puzzle. We are told that a jump from the $n=4$ state down to the $n=3$ state releases a photon in the **Ultraviolet (UV)** region of the electromagnetic spectrum. Our mission is to find which of the given transitions will produce a photon in the **Infrared (IR)** region.

### Decoding the Spectrum

To solve this, we must first recall the layout of the electromagnetic spectrum. The energy of a photon dictates its position on this spectrum. Ultraviolet light is highly energetic—it's the stuff that causes sunburns! Infrared light, on the other hand, is much lower in energy; we feel it as gentle heat.

Mathematically, we know that $E_{IR} < E_{Visible} < E_{UV}$. Therefore, the energy gap for the transition producing IR radiation must be **strictly less** than the energy gap of the $4 ightarrow 3$ transition that produced the UV radiation.

### The Master Equation

According to Bohr's model, the energy of an electron in the $n^{	ext{th}}$ orbit of a hydrogen-like atom is given by:

$$E_n = -\frac{13.6 Z^2}{n^2} 	ext{ eV}$$

When an electron transitions from an initial state $n_i$ to a final state $n_f$, the energy of the emitted photon is the difference between these two levels:

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

Since we are comparing transitions within the same atom, the $13.6 Z^2$ term is a constant. We only need to compare the proportionality factor: $\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}ight)$.

### Analyzing the Reference Transition

Let's calculate this factor for our reference UV transition, $4 ightarrow 3$:

$$\Delta E_{4 ightarrow 3} \propto \left(\frac{1}{3^2} - \frac{1}{4^2}ight)$$

$$\Delta E_{4 ightarrow 3} \propto \frac{1}{9} - \frac{1}{16} = \frac{16 - 9}{144} = \frac{7}{144} \approx 0.0486$$

Any transition that yields a factor **less than $0.0486$** will have lower energy and thus fall into the Infrared region.

### Testing the Options

Now, let's systematically test the given options:

**Option (a): $2 ightarrow 1$**
$$\Delta E_{2 ightarrow 1} \propto \left(\frac{1}{1^2} - \frac{1}{2^2}ight) = 1 - \frac{1}{4} = 0.75$$
This is massive! It's much larger than $0.0486$, meaning it's even higher energy than UV (likely extreme UV or X-rays).

**Option (b): $3 ightarrow 2$**
$$\Delta E_{3 ightarrow 2} \propto \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = \frac{1}{4} - \frac{1}{9} = \frac{5}{36} \approx 0.138$$
Still larger than our reference.

**Option (c): $4 ightarrow 2$**
$$\Delta E_{4 ightarrow 2} \propto \left(\frac{1}{2^2} - \frac{1}{4^2}ight) = \frac{1}{4} - \frac{1}{16} = \frac{3}{16} = 0.1875$$
Also larger.

**Option (d): $5 ightarrow 4$**
$$\Delta E_{5 ightarrow 4} \propto \left(\frac{1}{4^2} - \frac{1}{5^2}ight) = \frac{1}{16} - \frac{1}{25} = \frac{25 - 16}{400} = \frac{9}{400} = 0.0225$$

### The Grand Conclusion

Look at that! The factor for the $5 ightarrow 4$ transition is $0.0225$, which is **less than** the $0.0486$ of our UV transition.

Because $\Delta E_{5 ightarrow 4} < \Delta E_{4 ightarrow 3}$, the emitted photon has lower energy and a longer wavelength, placing it perfectly in the **Infrared** region. This beautifully aligns with the fact that transitions ending in $n=4$ belong to the Brackett series, which is known to be in the infrared spectrum.

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 radition will be obtained in the transition from

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Visualized Solution

$E n er g y L e v e l s$

$E M S p ec t r u m L o g i c$

$R e f er e n ce T r an s i t i o n$

$C h ec k in g O pt i o n (a)$

$C h ec k in g O pt i o n s (b) & (c)$

$C h ec k in g O pt i o n (d)$

$C o n c l u s i o n$

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

The Electromagnetic Puzzle

Decoding the Spectrum

The Master Equation

Analyzing the Reference Transition

Testing the Options

The Grand Conclusion

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