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Q18P

Expert-verifiedFound in: Page 1247

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**A hydrogen atom in its ground state actually has two possible, closely spaced energy levels because the electron is in the magnetic field $\overrightarrow{\mathbf{B}}$ of the proton (the nucleus). Accordingly, energy is associated with the orientation of the electron’s magnetic moment $\overrightarrow{\mathbf{\mu}}$ relative to $\overrightarrow{\mathbf{B}}$, and the electron is said to be either spin up (higher energy) or spin down (lower energy) in that field. If the electron is excited to the higher energy level, it can de-excite by spin-flipping and emitting a photon. The wavelength associated with that photon is 21 cm. (Such a process occurs extensively in the Milky Way galaxy, and reception of the 21 cm radiation by radio telescopes reveals where hydrogen gas lies between stars.) What is the effective magnitude of $\overrightarrow{\mathbf{B}}$ as experienced by the electron in the ground-state hydrogen atom?**

The effective magnitude of magnetic field $\overrightarrow{\mathrm{B}}$ as experienced by the electron in the ground-state hydrogen atom is 51 mT .

- The wavelength associated with that photon, $\lambda =0.21m$
- If the electron is excited to a higher level, it can de-excite by spin-flipping and emitting a photon.

**Magnetic resonance, absorption or radiation by electrons or atomic nuclei in response to the use of other magnetic fields.**

** **

**The power of attraction among all the crowds throughout the universe; especially the attraction of the gravity of the earth with the bodies near its surface.**

** **

**Use the concept of gravitational force to find gravitational potential energy. Integrate the equation of gravitational force over infinity to reference position. Then define the work required to increase the separation of the particles for ****the ****given positions.**

Formulas:

The energy equation from the Stern-Gerlach experiment,

$\u2206\mathrm{E}=2{\mathrm{\mu}}_{\mathrm{B}}\mathrm{B}$ ….. (1)

Here, B is the magnetic field and ${\mathrm{\mu}}_{\mathrm{B}}$ is the Bohr magneton.

The energy due to Planck-Einstein relation,

$\u2206\mathrm{E}=\frac{\mathrm{hc}}{\mathrm{\lambda}}$ ….. (2)

Here, c is the speed of light, $\lambda $ is the wavelength, and h is the Plank’s constant.** **

Consider the known data as below.

The wavelength, $\mathrm{\lambda}=21\mathrm{cm}=0.21\mathrm{m}$

Plank’s constant, $\mathrm{h}=6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}$

Speed of light, $\mathrm{c}=3\times {10}^{8}\mathrm{m}/\mathrm{s}$

Bohr magneton, ${\mathrm{\mu}}_{\mathrm{B}}=9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}$

Comparing equations (1) and (2) and substituting the given data in the derived equation of the magnetic field.

The value of the effective magnetic field is as follow.

$2{\mathrm{\mu}}_{\mathrm{B}}\mathrm{B}=\frac{\mathrm{hc}}{\mathrm{\lambda}}\phantom{\rule{0ex}{0ex}}\mathrm{B}=\frac{\mathrm{hc}}{\mathrm{\lambda}\left(2{\mathrm{\mu}}_{\mathrm{B}}\right)}$

Substitute known values in the above equation.

$\mathrm{B}=\frac{\left(6.63\times {10}^{-34}\mathrm{J}.\mathrm{s}\right)\left(3\times {10}^{8}\mathrm{m}/\mathrm{s}\right)}{2\left(0.21\mathrm{m}\right)\left(9.27\times {10}^{-24}\mathrm{J}/\mathrm{T}\right)}\phantom{\rule{0ex}{0ex}}=5.11\times {10}^{-2}\mathrm{T}\phantom{\rule{0ex}{0ex}}=51\mathrm{mT}$

Hence, the value of the magnetic field is 51 mT.

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