/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Is it possible for a hydrogen at... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 36: Problem 5

Is it possible for a hydrogen atom to be in the \(2 d\) state? Explain.

Short Answer

Expert verified
No, it is not possible for a hydrogen atom to be in the '2d' state. The quantum numbers of the atom do not allow for a '2d' state as the maximum possible \(l\) for \(n=2\) is 1, not 2.

Step by step solution

01

Understand the Quantum Numbers

The quantum representation of atoms consists of four quantum numbers: the energy level (principle quantum number, \(n\)), the angular momentum quantum number (\(l\)), the magnetic quantum number (\(m_l\)), and the spin quantum number (\(m_s\)). Each quantum number has its own set of valid values: \(n\) can be any positive integer, \(l\) ranges from 0 to \(n-1\), \(m_l\) ranges from \(-l\) to \(+l\), and \(m_s\) can be either +1/2 or -1/2.
02

Understanding the '2d' state

The provided state, '2d', implies that \(n=2\) and \(l\) corresponds to \(d\). Orbitals are commonly labelled as 's', 'p', 'd', and 'f' for \(l\) values 0, 1, 2, 3 respectively. Thus, for 'd', \(l=2\).
03

Verify the validity of the '2d' state

Given the quantum numbers, one can establish that for \(n=2\), the maximum possible value of \(l\) is \(2-1=1\) (or 'p'). Therefore, the '2d' state, which would correspond to \(n=2, l=2\), is not possible as \(l\) cannot equal \(n\) and must always be less than \(n\).

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.

Over 30 million students worldwide already upgrade their learning with 91影视!

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

Molybdenum's X-ray spectrum has its \(K \alpha\) peak at 17.4 keV. The corresponding X-ray wavelength is closest to a. \(1 \mathrm{pm}\) b. \(100 \mathrm{pm}\) c. \(1 \mathrm{nm}\) d. \(100 \mathrm{nm}\)

Singly ionized oxygen (so-called O-II) is a prevalent species in the tenuous gas between stars, and O-II emits a doublet spectral line at \(372.60 \mathrm{nm}\) and \(372.88 \mathrm{nm} .\) Astrophysicists analyze this line to learn, among other things, about the distribution of interstellar gas in distant galaxies. Find the energy splitting, in eV, that results in this doublet.

A hydrogen atom is in the \(2 s\) state. Find the probability that its electron will be found (a) beyond one Bohr radius and (b) beyond 10 Bohr radii.

The \(4 p \rightarrow 3\) s transition in sodium produces a double spectral line at \(330.237 \mathrm{nm}\) and \(330.298 \mathrm{nm} .\) What's the energy splitting of the \(4 p\) level?

Emission of characteristic \(\mathrm{X}\) rays occurs in the context of multielectron atoms that generally have all but one of their electrons present. You should therefore expect the X-ray energies to be described a. quite accurately by Bohr's atomic theory. b. through hydrogen-like solutions to the Schr枚dinger equation. c. only approximately by Bohr's theory or hydrogenic solutions to the Schr枚dinger equation.
See all solutions

Recommended explanations on Physics Textbooks

Electricity

Read Explanation

Torque and Rotational Motion

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Geometrical and Physical Optics

Read Explanation

Wave Optics

Read Explanation

Famous Physicists

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.