Electron spin: Radio astronomers can detect clouds of hydrogen too cool to radiate optical wavelengths of light by means of the 21 cm spectral line corresponding with the flipping of the electron in a hydrogen atom from having its spin parallel to the proton spin to having it antiparallel. From this wavelength, and thus E between states, find the magnetic field experienced by the electron in a hydrogen atom

Question

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nuhulawal20 nuhulawal20 · Answer 1 · 2021-04-16T12:13:01+02:00

Answer:

the magnetic field experienced by the electron is 0.0511 T

Explanation:

Given the data in the question;

Wavelength λ = 21 cm = 0.21 m

we know that Bohr magneton μ[tex]_B[/tex] is 9.27 × 10⁻²⁴ J/T

Plank's constant h is 6.626 × 10⁻³⁴ J.s

speed of light c = 3 × 10⁸ m/s

protein spin causes magnetic field in hydrogen atom.

so

Initial potential energy = -μ[tex]_B[/tex]B × cos0°

= -μ[tex]_B[/tex]B × 1

= -μ[tex]_B[/tex]B

Final potential energy = -μ[tex]_B[/tex]B × cos180°

= -μ[tex]_B[/tex]B × -1

= μ[tex]_B[/tex]B

so change in energy will be;

ΔE = μ[tex]_B[/tex]B - ( -μ[tex]_B[/tex]B )

ΔE = 2μ[tex]_B[/tex]B

now, difference in energy levels will be;

ΔE = hc/λ

2μ[tex]_B[/tex]B = hc/λ

2μ[tex]_B[/tex]Bλ = hc

B = hc / 2μ[tex]_B[/tex]λ

so we substitute

B = [(6.626 × 10⁻³⁴) × (3 × 10⁸)] / [2(9.27 × 10⁻²⁴) × 0.21 ]

B = [ 1.9878 × 10⁻²⁵ ] / [ 3.8934 × 10⁻²⁴ ]

B = 510556326.09

B = 0.0511 T

Therefore, the magnetic field experienced by the electron is 0.0511 T