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mamaluj [8]
3 years ago
6

Over the course of a martian year, what are the ranges of the rms speeds of the co2 molecules.

Physics
1 answer:
Nimfa-mama [501]3 years ago
3 0
<span>Speed (rms) = sqrt 3KT/m, where: K = Boltzmann's costant = 1.38*10^-23 joule/K; T = temperature in Kelvin degrees = 273 - 63 = 210 K; m = 44 * 1.672*10^-27 Kg; 1.672*10^-27 Kg (good approximation) is the weight of a single a.m.u. (atomic mass unit); 44 is molecular weight of CO2. Molecular Speed (CO2) = sqrt 3*1.38*10^-23*210/44*1.672*10^-27 = 344 m/sec. Excuse me for some imperfection in my language, I am from South Italy. My cordiality to You, hello.</span>
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The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron
FrozenT [24]

Answer:

<em>Speed of the electron is 2.46 x 10^8 m/s</em>

<em></em>

Explanation:

momentum of the electron before relativistic effect = M_{0} V

where M_{0} is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = M_{0}/ \sqrt{1 - \beta ^{2}  }

where

\beta = V/c

c  is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> M_{0}V/ \sqrt{1 - \beta ^{2}  }

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75M_{0} V = M_{0}V/ \sqrt{1 - \beta ^{2}  }

the equation reduces to

1.75 = 1/ \sqrt{1 - \beta ^{2}  }

square both sides of the equation, we have

3.0625 = 1/(1 - \beta ^{2} )

3.0625 - 3.0625\beta ^{2} = 1

2.0625 = 3.0625\beta ^{2}

\beta ^{2} = 0.67

β = 0.819

substitute for  \beta = V/c

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = <em>2.46 x 10^8 m/s</em>

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