Animated Solution for Physics - Atoms and Nuclei: Which of the following transitions in hydrogen atoms emit photons of highest frequency?

Imagine you are looking at the microscopic world of a hydrogen atom. It is a beautifully structured system, governed by the laws of quantum mechanics. The question asks us to identify which of the given electron transitions will emit a photon with the highest frequency.

This might seem like a straightforward calculation problem, but it is actually a profound conceptual test. It challenges our understanding of how electrons move between energy levels and how that movement translates into the light we can observe.

Before we dive into the math, let's visualize the setup. The hydrogen atom has distinct, quantized energy levels. The lowest energy level, closest to the nucleus, is the ground state ($n=1$). Above it are the excited states ($n=2,3,4,\dots$).

The first critical word in the question is "emit". What does it mean for an atom to emit a photon?

An electron in an atom can only exist in specific energy levels. When an electron absorbs energy, it jumps from a lower energy level to a higher energy level. This is called absorption.

Conversely, when an electron falls from a higher energy level to a lower energy level, it must release the excess energy. It does this by releasing a packet of light, a photon. This is called emission.

Therefore, for emission to occur, the initial principal quantum number ($n_i$) must be strictly greater than the final principal quantum number ($n_f$).

Let's look at our options with this filter: - Option (a): $n=2$ to $n=6$. The electron is moving to a higher level. This is absorption. - Option (d): $n=1$ to $n=2$. The electron is moving to a higher level. This is also absorption.

By simply understanding the physical meaning of the word "emit", we have instantly eliminated two out of the four options! We are now left with options (b) and (c), both of which represent emission.

Now we need to determine which of the remaining transitions emits a photon with the highest frequency.

To do this, we need to bridge the gap between the energy of the transition and the frequency of the emitted light. This bridge is provided by Max Planck's famous equation:

Here, $E$ is the energy of the photon, $h$ is Planck's constant, and $ u$ (nu) is the frequency of the photon.

Because the energy of the emitted photon is exactly equal to the energy difference between the two atomic levels ($\Delta E$), we can rewrite this as:

This equation tells us something incredibly important: frequency is directly proportional to the energy difference.

Therefore, to find the highest frequency, we simply need to find the transition that has the maximum energy difference. We don't need to calculate the actual frequency value; comparing the energy gaps is perfectly sufficient.

Let's calculate the energy difference for our two remaining candidates. The energy of an electron in the $n$-th orbit of a hydrogen atom is given by the Bohr formula:

Let's evaluate the energy levels involved in our options: - For $n=1$: $E_1=-\frac{13.6}{1^2}=-13.6\text{ eV}$ - For $n=2$: $E_2=-\frac{13.6}{2^2}=-3.4\text{ eV}$ - For $n=6$: $E_6=-\frac{13.6}{6^2}\approx -0.38\text{ eV}$

Now, let's calculate the energy released in each transition.

This transition belongs to the Balmer series. The energy difference is:

This transition belongs to the Lyman series. The energy difference is:

Let's compare the results. The energy released in the $n=2$ to $n=1$ transition ($10.2\text{ eV}$) is massively larger than the energy released in the $n=6$ to $n=2$ transition ($3.02\text{ eV}$).

Because $\Delta E_{2\to 1}>\Delta E_{6\to 2}$, the frequency of the photon emitted in the $n=2$ to $n=1$ transition will be significantly higher.

This reveals a beautiful and fundamental characteristic of the hydrogen atom's energy level structure. The energy levels are not evenly spaced. They get closer and closer together as $n$ increases.

The gap between the ground state ($n=1$) and the first excited state ($n=2$) is an enormous $10.2\text{ eV}$. This single gap is larger than the energy difference between $n=2$ and $n=\infty$ (which is only $3.4\text{ eV}$).

This means that any transition that ends at $n=1$ (the Lyman series) will always release more energy—and thus a higher frequency photon—than any transition that ends at $n=2$ (the Balmer series) or higher, regardless of where the electron starts from!

By mastering this conceptual shortcut, you can solve similar problems in seconds without doing any complex calculations. The transition dropping to the lowest possible $n$ will almost always yield the highest energy and highest frequency.

To truly appreciate the magnitude of these transitions, let's map them onto the electromagnetic spectrum. The energy of a photon dictates its position on the spectrum.

The transition from $n=6$ to $n=2$ releases $3.02\text{ eV}$ of energy. If we convert this energy into wavelength using the relation $\lambda =\frac{hc}{\Delta E}$, we find that this photon falls squarely within the visible light region. Specifically, it corresponds to the violet line in the hydrogen emission spectrum, a color you can see with your own eyes in a laboratory setting.

On the other hand, the transition from $n=2$ to $n=1$ releases a whopping $10.2\text{ eV}$ of energy. This massive energy packet corresponds to a much shorter wavelength, placing it deep within the ultraviolet (UV) region of the spectrum. This light is invisible to the human eye and carries enough energy to cause sunburns or even ionize other atoms.

This stark contrast highlights why the Lyman series (ending at $n=1$) is entirely in the ultraviolet region, while the Balmer series (ending at $n=2$) contains the visible lines. The sheer size of the energy gap between the ground state and the first excited state guarantees that any electron falling into the ground state will release a highly energetic, high-frequency UV photon.

When you encounter questions like this in competitive exams like JEE or NEET, time is of the essence. While calculating the exact energy values is a safe and rigorous approach, developing an intuitive feel for the energy level diagram is a superpower.

Always visualize the energy levels getting exponentially closer together as you move away from the nucleus. The "basement" drop (down to $n=1$) is always the steepest and most energetic fall. By keeping this visual anchor in your mind, you can confidently identify the highest frequency or shortest wavelength transitions in a matter of seconds, saving precious time for more complex calculations elsewhere in the paper.