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matrenka [14]
3 years ago
6

What is the difference in energy between a beta particle at rest and one traveling 0.35c?

Physics
1 answer:
monitta3 years ago
3 0

Answer:

They have a difference in energy of 35 eV.

Explanation:

The energy at rest of a particle is given by:

E_{R} = m_{0}c^2   (1)

Where m_{0} is the mass of the particle at rest and c is the speed of light.

Beta particles are high energy and high velocity electrons or positrons ejected from the nucleus of an atom as a consequence of a radioactive decay. Either if the beta particle is an electron¹ or a positron² it will have the same mass.

Hence, the mass of the beta particle at rest in equation (1) will be equal to the mass of an electron:

m_{e} = 9.1095x10^{-31} Kg

Replacing the values of m_{e} and c in equation (1) it is gotten:

E_{R} = (9.1095x10^{-31} Kg)(3.00x10^{8} m/s)^{2}

E_{R} = 8.19x10^{-14} Kg.m^{2}/s^{2}

But 1 J = Kg.m^{2}/s^{2}, therefore:

E_{R} = 8.19x10^{-14} J

It is better to express the rest energy in electronvolts (eV):

1eV = 1.60x10^{-19} J

8.19x10^{-14} J . \frac{1 eV}{1.60x10^{-19} J} ⇒ 511.875 eV

E_{R} = 511.875 eV

So the energy of the beta particle at rest is 511.875 eV.

Case for the one traveling at 0.35c:

Since it is traveling at 35% of the speed of light it is necessary to express equation (1) in a relativistic way, that can be done adding the Lorentz factor to it:

E = \frac{m_{0}c^{2}}{sqrt{1-\frac{v^{2}}{c^{2}}}}   (2)

Where v is the velocity of the particle (for this case 0.35c).

E = \frac{511.875 eV}{sqrt{1-\frac{(0.35c)^{2}}{c^{2}}}}

E = \frac{511.875 eV}{sqrt{1-\frac{0.1225c^{2}}{c^{2}}}}

E = \frac{511.875 eV}{sqrt{1-0.1225}}

E = \over{511.875 eV}{sqrt{0.8775}}

E = \over{511.875 eV}{0.936}

E = 546.875 eV

The difference in energy between the two particles can be determined using the relativistic form of the kinetic energy:

K = E – E_{R}  (3)

Where E is the energy of the particle traveling at 0.35c and E_{R} is the energy of the beta particle at rest.

K = 546.875 eV – 511.875 eV

K = 35 eV

They have a difference in energy of 35 eV.

Key terms:

¹Electron: Fundamental particle of negative electric charge.

²Positron: Is an electron with positive electric charge (similar to an electron in all its properties except in electric charge and magnetic moment).

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A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this
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Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

r=r_o\times A^{\frac{1}{3}}

r_o=1.25 \times 10^{-15} m = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

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r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}

r=4.6656\times 10^{-15} m=4.6656 fm

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) F=k\times \frac{q_1q_2}{a^2}

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q_1,q_2 = charges kept at distance 'a' from each other

F = electrostatic force between charges

q_1=+1.602\times 10^{-19} C

q_2=+1.602\times 10^{-19} C

Force of repulsion between two protons on opposite sides of the diameter

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F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}

F=2.6527 N

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