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soldier1979 [14.2K]
3 years ago
5

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
Physics
1 answer:
Ivan3 years ago
7 0

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_B 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_BB × cos0°

= -μ_BB × 1

= -μ_BB

Final potential energy = -μ_BB × cos180°

= -μ_BB × -1

= μ_BB

so change in energy will be;

ΔE = μ_BB - ( -μ_BB )

ΔE = 2μ_BB

now, difference in energy levels will be;

ΔE = hc/λ

2μ_BB = hc/λ

2μ_BBλ = hc

B = hc /  2μ_Bλ

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

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v = v_0 + at\ \to\ t = \frac{-v_0}{a} = \frac{-21.8\ m/s}{-2.4\ m/s^2} = \bf 9.08\ s

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Light waves have some similarities with water and sound waves, but they are not exactly the same. Describe all the differences y
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<u>Answer:</u>


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3 0
3 years ago
X rays of wavelength 0.0169 nm are directed in the positive direction of an x axis onto a target containing loosely bound electr
mamaluj [8]

Answer:

a) 4.04*10^-12m

b) 0.0209nm

c) 0.253MeV

Explanation:

The formula for Compton's scattering is given by:

\Delta \lambda=\lambda_f-\lambda_i=\frac{h}{m_oc}(1-cos\theta)

where h is the Planck's constant, m is the mass of the electron and c is the speed of light.

a) by replacing in the formula you obtain the Compton shift:

\Delta \lambda=\frac{6.62*10^{-34}Js}{(9.1*10^{-31}kg)(3*10^8m/s)}(1-cos132\°)=4.04*10^{-12}m

b) The change in photon energy is given by:

\Delta E=E_f-E_i=h\frac{c}{\lambda_f}-h\frac{c}{\lambda_i}=hc(\frac{1}{\lambda_f}-\frac{1}{\lambda_i})\\\\\lambda_f=4.04*10^{-12}m +\lambda_i=4.04*10^{-12}m+(0.0169*10^{-9}m)=2.09*10^{-11}m=0.0209nm

c) The electron Compton wavelength is 2.43 × 10-12 m. Hence you can use the Broglie's relation to compute the momentum of the electron and then the kinetic energy.

P=\frac{h}{\lambda_e}=\frac{6.62*10^{-34}Js}{2.43*10^{-12}m}=2.72*10^{-22}kgm\\

E_e=\frac{p^2}{2m_e}=\frac{(2.72*10^{-22}kgm)^2}{2(9.1*10^{-31}kg)}=4.06*10^{-14}J\\\\1J=6.242*10^{18}eV\\\\E_e=4.06*10^{-14}(6.242*10^{18}eV)=0.253MeV

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4 years ago
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