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DedPeter [7]
4 years ago
6

X-ray photons of wavelength 0.02480 nm are incident on a target and the Compton scattered photons are observed at 90 degrees.

Physics
1 answer:
Rashid [163]4 years ago
8 0

Answer:

Explanation:

(a) This problem can be solved by using the formula for Compton effect

\lambda_{a}-\lambda_{b}=\frac{h}{m_{o}c}(1-cos\phi)

where h is the planck constant, Φ is the angle of the scattered photon, m is the mass of the particle (in this case we take an electron) and c is the speed of ligth.

By taking the wavelength of the scattered photon and by replacing we have

\lambda_{a}=\frac{6.62*10^{-34}m^{2}\frac{kg}{s}}{(9.1*10^{-31}kg)(3*10^{8}\frac{m}{s})}+0.02480*10^{-9}m=2.72*10^{-11}m=0.0272nm

(b)

p=\frac{h}{\lambda_{b}}=\frac{6.62*10^{-34}m^{2}\frac{kg}{s}}{0.0248*10^{-9}m}=2.66*10^{-23}kg\frac{m}{s}

(c)

The energy is conserved, hence we have

E_{i-photon}=E_{k-electron}+E_{s-photon}

that is, the sum of the kinetic energy of the scattered electron and the energy of the scattered photon is equal to the energy of the incident photon. By taking Ek we have

E_{k-e}=E_{i-p}-E_{s-p}=h\frac{c}{\lambda_{b}}-h\frac{c}{\lambda_{a}}\\\\E_{k-e}=(6.62*10^{-34}m^{2}\frac{kg}{s})(3*10^{8}\frac{m}{s})(\frac{1}{0.0248*10^{-9}m}-\frac{1}{0.0272*10^{-9}m})\\\\E_{k-e}=7.06*10^{-16}J=7.06*10^{-16}(6.242*10^{18}eV)=4.4keV

(d)

E_{k-e}=\frac{m_{e}v^{2}}{2}\\\\v=\sqrt{\frac{2E_{k-e}}{m_{e}}}=3.9*10^{7}\frac{m}{s}\\\\p_{e}=m_{e}*v=3.5*10^{-23}kg\frac{m}{s}

I hope this is useful for you

regards

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Answer:

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Explanation:

Given the data in the question;

Using the Clapeyron equation

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(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

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Free runners jump long distances and land on the ground or a wall. How do they apply Newton’s second law to lessen the force of
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The potential difference,electric current ,resistance and new electric current will be 12 V,4 A,3 Ω,2 A.

<h3>What is resistance?</h3>

Resistance is a type of opposition force due to which the flow of current is reduced in the material or wire. Resistance is the enemy of the flow of current.

The energy in terms of the charge and potential difference is;

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Explanation:

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