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vodomira [7]
3 years ago
10

: match each description below with the type of star that is described. You may assume that all of the stars described have appr

oximately the same mass as the Sun. This is an adaptation of Thought Question 16, page 847 in OpenStax Astronomy. A. A star with no nuclear reactions going on in the core, which is made primarily of carbon and oxygen B. A star of uniform composition from center to surface; it contains hydrogen but has no nuclear reactions going on in the core. C. A star that is fusing hydrogen to form helium in its core. D. A star that is fusing helium to carbon in the core and hydrogen to helium in a shell around the core.
Physics
1 answer:
EleoNora [17]3 years ago
5 0

Explanation:

A. White Dwarf: It is the end stage of a Sun-like star. At the end the red giant star will throw away the outer layers in the form a shell and in the center a hot and bright core will remain. This core consists of carbon and oxygen but the conditions are not suitable to start the fusion of carbon and oxygen.

B. Protostar: It is the initial stage of a star. Here the fuel is available but still the temperature and pressure are not enough to start the nuclear fusion.

C. Main Sequence Star: A star that is fusing Hydrogen in the core to make Helium and releasing the energy in this process. Sun is currently in its main sequence stage. A star spend 90% of its lifetime in main sequence stage.

D. Red Giant Star: In such stars, the hydrogen has been exhausted and only Helium is left. As the core starts contracting under gravity, the conditions are apt for the fusion of Helium to begin. Helium starts fusing to make carbon while in the outer layer Hydrogen to Helium process goes on. As a result the star expands.

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Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

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

here, \lambda is the initial photon wavelength, \lambda^{'} is the scattered photon wavelength, h is he Planck's constant, m_e is the free electron mass, c is the velocity of light, \theta is the angle of scattering.

Given that, the scattering angle is, \theta=147^{\circ}

Putting the respective values, we get

\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.

Here, the photon's incident wavelength is \lamda=2.78pm

Therefore,

\lambda^{'}=2.78+4.45=7.23 pm

From the conservation of momentum,

\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}

where,\vec{P_\lambda} is the initial photon momentum, \vec{P_{\lambda^{'}}} is the final photon momentum and \vec{P_e} is the scattered electron momentum.

Expanding the vector sum, we get

P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta

Now expressing the momentum in terms of De-Broglie wavelength

P=h/\lambda,

and putting it in the above equation we get,

\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}

Therefore,

\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm

This is the de Broglie wavelength of the electron after scattering.

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