An excited hydrogen atom, whose electron is in an \(n=4\) state, is motionless. When the electron falls into the ground state, does it set the atom in motion? If so, with what speed?

Short Answer

Expert verified
A: Yes, the hydrogen atom starts moving, and its speed is approximately \(3.63\times 10^{-2}\,\text{m/s}\).

Step by step solution

01

Calculate the energy released during the transition

The energy released when an electron falls from an initial state \(n_i\) to a final state \(n_f\) can be calculated using the following formula: \(E = -13.6\,\text{eV}\left(\frac{1}{n_f^2} - \frac{1}{n_i^2}\right)\) In our case, the initial state \(n_i = 4\) and the final state (ground state) \(n_f = 1\). Plugging these values into the formula gives: \(E = -13.6\,\text{eV}\left(\frac{1}{1^2} - \frac{1}{4^2}\right) = -13.6\,\text{eV}\left(1 - \frac{1}{16}\right)\) \(E = -13.6\,\text{eV}\left(\frac{15}{16}\right) = -12.75\,\text{eV}\) The energy released during the transition is \(12.75\,\text{eV}\).
02

Convert the energy to Joules

To calculate the speed, we need to convert the energy from electron volts (eV) to Joules (J). The conversion factor is \(1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}\). Therefore, \(E = -12.75\,\text{eV}\times 1.6\times 10^{-19}\,\text{J/eV} = -2.04\times 10^{-18}\,\text{J}\).
03

Calculate the velocity of the electron

When the electron drops from the n=4 state to the ground state, it gains kinetic energy equal to the energy released during the transition. The conservation of momentum dictates that the hydrogen atom (proton) gains an equal amount of momentum but in the opposite direction. We can use the following equation to calculate the velocity of the electron: \(v_e = \sqrt{\frac{2E}{m_e}}\) where \(m_e\) is the mass of the electron (\(9.11\times10^{-31}\,\text{kg}\)). \(v_e = \sqrt{\frac{2\times -2.04\times 10^{-18}\,\text{J}}{9.11\times10^{-31}\,\text{kg}}} = 6.674\times 10^5\,\text{m/s}\)
04

Calculate the hydrogen atom's velocity

To calculate the velocity of the hydrogen atom, we can use the conservation of momentum. The mass of the proton (\(m_p\)) is \(1.67\times 10^{-27}\,\text{kg}\). The total initial momentum is zero. \(m_p v_p + m_e v_e = 0\) Solving for the proton's velocity (\(v_p\)): \(v_p = -\frac{m_e v_e}{m_p} = -\frac{(9.11\times10^{-31}\,\text{kg})(6.674\times 10^5\,\text{m/s})}{1.67\times 10^{-27}\,\text{kg}}\) \(v_p = 3.628\times 10^{-2}\,\text{m/s}\) The hydrogen atom will be set in motion, and its speed will be approximately \(3.63\times 10^{-2}\,\text{m/s}\).

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 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

The muon has the same charge as an electron but a mass that is 207 times greater. The negatively charged muon can bind to a proton to form a new type of hydrogen atom. How does the binding energy \(E_{\mathrm{B} \mu}\) of the muon in the ground state of a muonic hydrogen atom compare with the binding energy \(E_{\mathrm{Be}}\) of an electron in the ground state of a conventional hydrogen atom? a) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right|\) d) \(\left|E_{\mathrm{B} \mu}\right| \approx 200 \mid E_{\mathrm{Be}}\) b) \(\left|E_{\mathrm{B} \mu}\right| \approx 100\left|E_{\mathrm{Be}}\right|\) e) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 200\) c) \(\left|E_{\mathrm{B} \mu}\right| \approx\left|E_{\mathrm{Be}}\right| / 100\)

Section 38.2 established that an electron, if observed in the ground state of hydrogen, would be expected to have an observed speed of \(0.0073 c .\) For what atomic charge \(Z\) would an innermost electron have a speed of approximately \(0.500 c,\) when considered classically?

What is the energy of the orbiting electron in a hydrogen atom with a quantum number of \(45 ?\)

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

By what percentage is the electron mass changed in using the reduced mass for the hydrogen atom? What would the reduced mass be if the proton had the same mass as the electron?

See all solutions

Recommended explanations on Physics Textbooks

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