Consider a hydrogen atom with v_k, r_k and K_k denoting the velocity, orbital radius and kinetic ene

Q: Why is the change in total energy equal to the change in kinetic energy?

In a hydrogen atom, the potential energy $U$ is $-2K$, so the total energy $E = K + U = -K$. Therefore, the change in total energy $\Delta E = -\Delta K$, and their magnitudes are exactly equal.

Q: Why did we eliminate mass in option A?

The expression given in option A does not contain the mass term $m$. To verify if it's correct, we must use Bohr's quantization condition to substitute $m$ in terms of $v$, $r$, and $n$.

Q: Is the Lyman series always ending at $n=1$?

Yes! By definition, the Lyman series corresponds to electron transitions from any higher energy level ($n > 1$) down to the ground state ($n=1$).

Q: Why doesn't option B work even though it has kinetic energy in the denominator?

The de Broglie wavelength is inversely proportional to the *square root* of kinetic energy ($\lambda \propto 1/\sqrt{K}$). Option B provides an expression with the difference of inverse kinetic energies ($1/K$), which is mathematically different from the difference of their square roots.

Q: Can we just memorize these formulas instead of deriving them?

While you can memorize the standard formulas for $v_n$, $r_n$, and $E_n$, questions in JEE Advanced often test your ability to manipulate the fundamental postulates (like $mvr = nh/2\pi$) to create novel expressions. Deriving them builds the required analytical muscle.

Question

Consider a hydrogen atom with $v_k$, $r_k$ and $K_k$ denoting the velocity, orbital radius and kinetic energy of the electron in the $k^{	ext{th}}$ orbit, respectively. The electron undergoes a transition from the $n^{	ext{th}}$ orbit, emitting radiation corresponding to the Lyman series. Considering $h$ to be the Planck's constant and $\epsilon_0$ the permittivity of the free space, the correct statement(s) is/are:

Options:

(A) Magnitude of change in kinetic energy of electron can be expressed as $\frac{h}{4\pi}\left|\frac{nv_n}{r_n} - \frac{v_1}{r_1}ight|$.

(B) Magnitude of change in de Broglie wavelength of the electron can be expressed as $\frac{e^2}{4\epsilon_0}\left|\frac{1}{K_n} - \frac{1}{K_1}ight|$.

(C) Frequency of the radiation emitted can be expressed as $\frac{e^2}{8\pi\epsilon_0 h}\left(\frac{1}{r_1} - \frac{1}{r_n}ight)$.

(D) Magnitude of change in total energy of the electron can be expressed as $\frac{h}{2\pi}\left|\frac{v_1}{r_1} - \frac{nv_n}{r_n}ight|$.

Sigma Percentile · Accepted Answer

### The Quantum Leap: Unraveling the Hydrogen Atom

Imagine you are shrinking down to the subatomic realm, standing right next to a hydrogen atom. At the center lies a single, positively charged proton, and orbiting it is a lone electron. But this isn't a chaotic swarm; it's a highly structured, quantized dance governed by the elegant rules of the Bohr model.

In this problem, we are witnessing a classic quantum event: the **Lyman series transition**. Our electron is taking a leap of faith from a higher energy orbit $n$ down to the ground state, where $n=1$. As it plunges into this lower energy state, it sheds its excess energy by firing off a photon. Our mission is to mathematically decode the changes in its kinetic energy, de Broglie wavelength, frequency, and total energy during this dramatic fall.

### Analyzing Option A: The Kinetic Energy Puzzle

Let's start with the kinetic energy, $K$. We know the classical formula is $K = \frac{1}{2}mv^2$. However, the expression in Option A doesn't have the mass $m$ anywhere! How do we bridge this gap?

This is where Niels Bohr's stroke of genius comes in: the **quantization of angular momentum**. Bohr postulated that the angular momentum $mvr$ must be an integral multiple of $\frac{h}{2\pi}$. For the $k^{	ext{th}}$ orbit, this is written as:
$$m v_k r_k = \frac{k h}{2\pi}$$

We can cleverly rearrange this to isolate the mass:
$$m = \frac{k h}{2\pi v_k r_k}$$

Now, let's substitute this quantum mass back into our classical kinetic energy equation:
$$K_k = \frac{1}{2} \left( \frac{k h}{2\pi v_k r_k} ight) v_k^2 = \frac{k h v_k}{4\pi r_k}$$

Notice how beautifully one of the velocity terms cancels out! To find the magnitude of the change in kinetic energy during the transition from $n$ to $1$, we simply take the absolute difference:
$$\Delta K = |K_n - K_1| = \left| \frac{n h v_n}{4\pi r_n} - \frac{1 \cdot h v_1}{4\pi r_1} ight|$$

Factoring out the common $\frac{h}{4\pi}$, we get:
$$\Delta K = \frac{h}{4\pi} \left| \frac{n v_n}{r_n} - \frac{v_1}{r_1} ight|$$

This perfectly matches Option A. The math doesn't lie; Option A is correct!

### Analyzing Option B: The Wavelength Trap

Next, we evaluate the change in the electron's de Broglie wavelength. The de Broglie wavelength $\lambda$ is intimately tied to momentum $p$ by the relation $\lambda = \frac{h}{p}$. Since kinetic energy $K = \frac{p^2}{2m}$, we can express momentum as $p = \sqrt{2mK}$.

Substituting this into the wavelength formula gives:
$$\lambda_k = \frac{h}{\sqrt{2m K_k}}$$

The change in wavelength is the difference between the initial and final states:
$$|\lambda_n - \lambda_1| = \frac{h}{\sqrt{2m}} \left| \frac{1}{\sqrt{K_n}} - \frac{1}{\sqrt{K_1}} ight|$$

Look closely at this result. The change involves the difference of the **inverse square roots** of the kinetic energies. Option B, however, suggests an expression proportional to $\left| \frac{1}{K_n} - \frac{1}{K_1} ight|$, which is the difference of the inverse kinetic energies themselves. These are fundamentally different mathematical structures. Therefore, Option B is a cleverly disguised trap and is incorrect.

### Analyzing Option C: The Photon's Frequency

When the electron drops to a lower orbit, the energy it loses is carried away by a single photon. According to Planck's equation, the energy of this photon is $E_{	ext{photon}} = hf$, where $f$ is the frequency.

By conservation of energy, the photon's energy equals the difference in the electron's total energy:
$$hf = E_n - E_1$$

In the Bohr model, the total energy $E$ is exactly the negative of the kinetic energy $K$. So, $E_n - E_1 = -K_n - (-K_1) = K_1 - K_n$. 
We also know the standard expression for kinetic energy in terms of the orbital radius:
$$K_k = \frac{e^2}{8\pi\epsilon_0 r_k}$$

Substituting this into our frequency equation:
$$f = \frac{K_1 - K_n}{h} = \frac{1}{h} \left( \frac{e^2}{8\pi\epsilon_0 r_1} - \frac{e^2}{8\pi\epsilon_0 r_n} ight)$$

Factoring out the constants, we arrive at:
$$f = \frac{e^2}{8\pi\epsilon_0 h} \left( \frac{1}{r_1} - \frac{1}{r_n} ight)$$

This is an exact match for Option C. The physics holds up perfectly!

### Analyzing Option D: The Total Energy Check

Finally, let's look at the change in total energy. As we established earlier, the total energy $E$ is the negative of the kinetic energy $K$. Therefore, the magnitude of the change in total energy is identical to the magnitude of the change in kinetic energy:
$$|\Delta E| = |\Delta K|$$

From our work on Option A, we already know that:
$$|\Delta K| = \frac{h}{4\pi} \left| \frac{v_1}{r_1} - \frac{n v_n}{r_n} ight|$$

Option D presents an almost identical expression, but with a crucial flaw: it uses a prefactor of $\frac{h}{2\pi}$ instead of $\frac{h}{4\pi}$. It is off by a factor of two! Thus, Option D is incorrect.

### The Final Verdict

By systematically applying the core postulates of the Bohr model and carefully navigating the algebraic manipulations, we have successfully decoded the problem. The correct statements are **(A)** and **(C)**. This journey highlights the profound interconnectedness of velocity, radius, and energy in the quantum world.

Select Answer:

Visualized Solution

Lyman Series Transition: $n \to 1$

Kinetic Energy \& Quantization

Eliminating Mass $m$

Evaluating Option A

de Broglie Wavelength

Evaluating Option B

Frequency of Emitted Photon

Evaluating Option C

Evaluating Option D

Final Conclusion

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

The Quantum Leap

Analyzing Option A

Analyzing Option B

Analyzing Option C

Analyzing Option D

The Final Verdict

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