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____ [38]
4 years ago
12

Bowen’s reaction series illustrates relations between:

Physics
1 answer:
Finger [1]4 years ago
5 0

C. Temperature, chemical composition and mineral structure

Explanation:

The Bowen's reaction series illustrates the relationship between temperature, chemical composition and mineral structure.

The series is made up of a continuous and discontinuous end through which magmatic composition can be understood as temperature changes.

  • The left part is the discontinuous end while the right side is the continuous series.
  • From the series, we understand that a magmatic body becomes felsic as it begins to cool to lower temperature.
  • A magma at high temperature is ultramafic and very rich in ferro-magnesian silicates which are the chief mineral composition of olivine and pyroxene. These minerals are predominantly found in mafic- ultramafic rocks. Also, we expect to find the calcic-plagioclase at high temperatures partitioned in the magma.
  • At a relatively low temperature, minerals with frame work structures begins to form . The magma is more enriched with felsic minerals and late stage crystallization occurs here.

Learn more:

Silicate minerals brainly.com/question/4772323

#learnwithBrainly

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X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
Calculate the energy of the green light emitted, per photon, by a mercury lamp with a frequency of 5.49 × 1014 hz.
Tcecarenko [31]
The energy of a photon is given by
E=hf
where
h=6.6 \cdot 10^{-34} Js is the Planck constant
f is the frequency of the photon

In our problem, the frequency of the light is 
f=5.49 \cdot 10^{14}Hz
therefore we can use the previous equation to calculate the energy of each photon of the green light emitted by the lamp:
E=hf=(6.6 \cdot 10^{-34}Js)(5.49 \cdot 10^{14} Hz)=3.62 \cdot 10^{-19} J
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A cyclist moves at a constant speed of 5 m/s if the cyclist does not accelerate during the next 20 seconds he will travel at?
Verizon [17]
5 m/s because the speed is constant 
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3 years ago
Compare skater's total energy at point A and at point E?(disregard the friction)
White raven [17]

Answer:

c

Explanation:

because I've had this question before and got it right

7 0
3 years ago
What best describes the significance of albert einstein's famous equation famous equations e=mc2?
babunello [35]
The absolute simplest way to explain E=MC^2 is: a lot of energy from a little mass.
8 0
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