Wavelength Of Photon Emission In Hydrogen Atom Transition

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Hey guys! Today, let's dive into a fascinating topic in physics: calculating the wavelength of a photon emitted when a hydrogen atom transitions between energy levels. Specifically, we're tackling the question of what the wavelength is when an electron jumps from the ni=3{ n_i = 3 } energy level down to the nf=2{ n_f = 2 } level. This is a classic problem in atomic physics, and understanding it will give you a solid grasp of how atoms emit light.

The Rydberg Formula: Our Key Tool

To figure this out, we'll need to use a powerful tool called the Rydberg formula. This formula is your best friend when dealing with electron transitions and photon wavelengths in hydrogen atoms. It looks like this:

1λ=R(1nf2−1ni2){ \frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) }

Where:

  • λ{ \lambda } is the wavelength of the emitted photon (what we want to find!).
  • R{ R } is the Rydberg constant, which is approximately 1.097×107 m−1{ 1.097 \times 10^7 \text{ m}^{-1} }. This constant is a fundamental value in atomic physics, derived from experimental measurements and the Bohr model of the atom.
  • ni{ n_i } is the initial energy level (the level the electron starts at).
  • nf{ n_f } is the final energy level (the level the electron ends up at).

Breaking it Down:

Let's break down why this formula works. The Rydberg formula is essentially a mathematical expression of the energy differences between electron energy levels in a hydrogen atom. When an electron transitions from a higher energy level to a lower one, it releases energy in the form of a photon. The energy of this photon is directly related to its wavelength, and the Rydberg formula neatly ties together the initial and final energy levels with the resulting wavelength.

The Significance of the Rydberg Constant:

The Rydberg constant (R{ R }) is crucial. It embodies the fundamental properties of the hydrogen atom and the nature of electromagnetic radiation. Its value is derived from a combination of other fundamental constants, such as the Planck constant, the speed of light, and the elementary charge. This highlights how interconnected different areas of physics are, and how a single constant can encapsulate so much information.

Why This Formula Matters:

This formula is not just some abstract equation; it has profound implications. It allows us to predict the specific wavelengths of light that hydrogen atoms will emit under different conditions. This is incredibly useful in fields like spectroscopy, where we analyze the light emitted by substances to determine their composition and properties. For example, astronomers use this principle to study the light from distant stars and galaxies, allowing them to understand the elements present in those celestial objects.

Plugging in the Values: Solving for the Wavelength

Now, let's get our hands dirty and plug in the values for our specific problem. We know that:

  • ni=3{ n_i = 3 } (initial level)
  • nf=2{ n_f = 2 } (final level)
  • R=1.097×107 m−1{ R = 1.097 \times 10^7 \text{ m}^{-1} } (Rydberg constant)

Let's substitute these values into the Rydberg formula:

1λ=(1.097×107 m−1)(122−132){ \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \left( \frac{1}{2^2} - \frac{1}{3^2} \right) }

Step-by-Step Calculation:

  1. Calculate the fractions inside the parentheses: 122=14=0.25{ \frac{1}{2^2} = \frac{1}{4} = 0.25 } 132=19≈0.111{ \frac{1}{3^2} = \frac{1}{9} \approx 0.111 }
  2. Subtract the fractions: 0.25−0.111=0.139{ 0. 25 - 0.111 = 0.139 }
  3. Multiply by the Rydberg constant: 1λ=(1.097×107 m−1)×0.139≈1.525×106 m−1{ \frac{1}{\lambda} = (1.097 \times 10^7 \text{ m}^{-1}) \times 0.139 \approx 1.525 \times 10^6 \text{ m}^{-1} }

Finding the Wavelength:

We've calculated the reciprocal of the wavelength, so now we need to take the inverse to find the actual wavelength:

λ=11.525×106 m−1≈6.55×10−7 m{ \lambda = \frac{1}{1.525 \times 10^6 \text{ m}^{-1}} \approx 6.55 \times 10^{-7} \text{ m} }

Converting to Angstroms: Getting the Right Units

The question asks for the wavelength in Angstroms (Å). Remember that 1 Å is equal to 10−10{ 10^{-10} } meters. So, to convert meters to Angstroms, we need to multiply by 1010{ 10^{10} }:

λ (in A˚)=(6.55×10−7 m)×(1010 A˚/m)=6550 A˚{ \lambda \text{ (in Å)} = (6.55 \times 10^{-7} \text{ m}) \times (10^{10} \text{ Å/m}) = 6550 \text{ Å} }

Therefore, the wavelength of the photon emitted during this transition is approximately 6550 Å.

Units Matter!

It's crucial to pay attention to units in physics problems. Always make sure your units are consistent and that you're converting to the correct units as needed. In this case, the problem specifically asked for the answer in Angstroms, so we had to perform that final conversion step.

Connecting to the Balmer Series: A Little Extra Knowledge

This particular transition, from ni=3{ n_i = 3 } to nf=2{ n_f = 2 }, is part of what's known as the Balmer series. The Balmer series includes all transitions in the hydrogen atom where the electron ends up in the n=2{ n = 2 } energy level. These transitions result in the emission of photons in the visible and ultraviolet portions of the electromagnetic spectrum. The specific wavelength we calculated (6550 Å) corresponds to a red light, which is a prominent line in the hydrogen emission spectrum.

Other Series:

There are other series as well, such as the Lyman series (transitions to n=1{ n = 1 }, ultraviolet light) and the Paschen series (transitions to n=3{ n = 3 }, infrared light). Each series corresponds to a different range of wavelengths in the electromagnetic spectrum, and they provide a wealth of information about the energy level structure of the hydrogen atom.

So, What's the Answer?

Looking back at the original multiple-choice options, the closest answer to our calculated value of 6550 Å is:

  • a. 6.56×103{ 6.56 \times 10^3 } Å

Therefore, option a is the correct answer.

Key Takeaways and Pro Tips

Let's wrap up with some key takeaways and tips for tackling similar problems:

  • Master the Rydberg Formula: This formula is your bread and butter for hydrogen atom transitions. Memorize it and understand what each term represents.
  • Pay Attention to Units: Always be mindful of units and make sure to convert them appropriately.
  • Step-by-Step Approach: Break down complex calculations into smaller, manageable steps. This reduces the chance of errors.
  • Understand the Physics: Don't just memorize formulas; strive to understand the underlying physics concepts. This will help you apply the formulas correctly and solve a wider range of problems.
  • Practice, Practice, Practice: The more you practice, the more comfortable you'll become with these types of calculations.

Pro Tips for Exam Success:

  • Double-Check Your Work: In exams, time is precious, but it's always worth spending a few extra seconds to double-check your calculations.
  • Estimate Your Answer: Before diving into calculations, try to estimate the order of magnitude of the answer. This can help you spot errors in your calculations.
  • Show Your Work: Even if you make a mistake, showing your work can earn you partial credit.

Wrapping Up: You've Got This!

Calculating photon wavelengths in hydrogen atom transitions might seem daunting at first, but with the Rydberg formula and a step-by-step approach, you can master it. Remember to understand the underlying physics, pay attention to units, and practice consistently. You guys have got this! Keep exploring the fascinating world of quantum physics!

If you have any more questions or want to dive deeper into atomic physics, just let me know. Happy calculating!