Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 38: Problem 49

What is the shortest possible wavelength of the Lyman series in hydrogen?

Short Answer

Expert verified
Answer: The shortest possible wavelength of the Lyman series in hydrogen is approximately 91.1 nm.

Step by step solution

01

Recall the formula for the hydrogen spectral lines

To calculate the wavelength of the spectral lines in hydrogen, we will use the Rydberg formula for hydrogen: $$\frac{1}{\lambda} = R_H \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ In this formula, λ represents the wavelength, 𝑅_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m⁻¹), n₁ and n₂ are the principal quantum numbers of the initial and final energy levels, respectively.
02

Determine n₁ and n₂ for the Lyman series

For the Lyman series, electron transitions occur between the first energy level (n = 1) and any higher energy level (n > 1). To find the shortest wavelength, we need the transition with the highest energy difference, which corresponds to an electron transitioning from n -> infinity (n₂ = ∞) to the first energy level (n₁ = 1).
03

Plug the values into the Rydberg formula

Now that we have n₁ = 1 and n₂ = ∞, we can plug these values into the Rydberg formula: $$\frac{1}{\lambda} = R_H \left(\frac{1}{1^2} - \frac{1}{\infty^2}\right)$$
04

Simplify the equation

Now, simplifying the equation, we get: $$\frac{1}{\lambda} = R_H (1 - 0)$$ $$\frac{1}{\lambda} = R_H$$
05

Find the shortest possible wavelength

Now, we can solve for λ by taking the reciprocal of both sides of the equation: $$\lambda = \frac{1}{R_H}$$ Plugging in the value of 𝑅_H (1.097 x 10^7 m⁻¹): $$\lambda = \frac{1}{1.097 \times 10^7 \mathrm{m^{-1}}}$$ Thus, the shortest possible wavelength of the Lyman series in hydrogen is: $$\lambda ≈ 9.11 \times 10^{-8} \mathrm{m}$$ or approximately 91.1 nm.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following can be used to explain why you can't walk through walls? a) Coulomb repulsion d) the Pauli exclusion b) the strong nuclear force \(\quad\) principle c) gravity e) none of the above

The wavelength of the fourth line in the Lyman series is a) \(80.0 \mathrm{nm}\). b) \(85.0 \mathrm{nm}\). c) \(90.2 \mathrm{nm}\). d) \(94.9 \mathrm{nm}\).

A muon is a particle very similar to an electron. It has the same charge but its mass is \(1.88 \cdot 10^{-28} \mathrm{~kg}\). a) Calculate the reduced mass for a hydrogen-like muonic atom consisting of a single proton and a muon. b) Calculate the ionization energy for such an atom, assuming the muon starts off in its ground state.

A collection of hydrogen atoms have all been placed into the \(n=4\) excited state. What wavelengths of photons will be emitted by the hydrogen atoms as they transition back to the ground state?

An excited hydrogen atom emits a photon with an energy of \(1.133 \mathrm{eV}\). What were the initial and final states of the hydrogen atom before and after emitting the photon?
See all solutions

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Read Explanation

Electric Charge Field and Potential

Read Explanation

Atoms and Radioactivity

Read Explanation

Modern Physics

Read Explanation

Thermodynamics

Read Explanation

Electrostatics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free