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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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Fan object moves in uniform circular motion in a circle of radius R=200 meters, and the objectes 5.00 seconds to
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The centripetal acceleration of the object is 31550.72\ m/s^2.  

Explanation:

We have,

Radius of a circular path is 200 m

It takes 5 seconds to complete 10 revolutions. The angular velocity of the object is given by the rate of change of angular displacement per unit time :

\omega=\dfrac{\Delta \theta}{\Delta t}\\\\\omega=\dfrac{2\pi \times 10}{5}\\\\\omega=12.56\ rad/s

The centripetal acceleration of the object is given by :

a=\omega^2 r\\\\a=(12.56)^2 \times 200\\\\a=31550.72\ m/s^2

So, the centripetal acceleration of the object is 31550.72\ m/s^2.

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Two parallel plates of area 0.155 m2<br> are separated by 0.00100 m. What<br> is their capacitance?
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The British gold sovereign coin is an alloy of gold and copper having a total mass of 7.988 g, and is 22-karat gold 24 x (mass o
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Answers:

(a) 0.0073kg

(b) Volume gold: 3.79(10)^{-7}m^{3}, Volume cupper: 7.6(10)^{-8}m^{3}

(c) 17633.554kg/m^{3}

Explanation:

<h2>(a) Mass of gold </h2><h2 />

We are told the total mass M of the coin, which is an alloy  of gold and copper is:

M=m_{gold}+m_{copper}=7.988g=0.007988kg   (1)

Where  m_{gold} is the mass of gold and m_{copper} is the mass of copper.

In addition we know it is a 22-karat gold and the relation between the number of karats K and mass is:

K=24\frac{m_{gold}}{M}   (2)

Finding {m_{gold}:

m_{gold}=\frac{22}{24}M   (3)

m_{gold}=\frac{22}{24}(0.007988kg)   (4)

m_{gold}=0.0073kg   (5)  This is the mass of gold in the coin

<h2>(b) Volume of gold and cupper</h2><h2 />

The density \rho of an object is given by:

\rho=\frac{mass}{volume}

If we want to find the volume, this expression changes to: volume=\frac{mass}{\rho}

For gold, its volume V_{gold} will be a relation between its mass m_{gold}  (found in (5)) and its density \rho_{gold}=19.30g/cm^{3}=19300kg/m^{3}:

V_{gold}=\frac{m_{gold}}{\rho_{gold}}   (6)

V_{gold}=\frac{0.0073kg}{19300kg/m^{3}}   (7)

V_{gold}=3.79(10)^{-7}m^{3}   (8)  Volume of gold in the coin

For copper, its volume V_{copper} will be a relation between its mass m_{copper}  and its density \rho_{copper}=8.96g/cm^{3}=8960kg/m^{3}:

V_{copper}=\frac{m_{copper}}{\rho_{copper}}   (9)

The mass of copper can be found by isolating m_{copper} from (1):

M=m_{gold}+m_{copper}  

m_{copper}=M-m_{gold}  (10)

Knowing the mass of gold found in (5):

m_{copper}=0.007988kg-0.0073kg=0.000688kg  (11)

Now we can find the volume of copper:

V_{copper}=\frac{0.000688kg}{8960kg/m^{3}}   (12)

V_{copper}=7.6(10)^{-8}m^{3}   (13)  Volume of copper in the coin

<h2>(c) Density of the sovereign coin</h2><h2 />

Remembering density is a relation between mass and volume, in the case of the coin the density \rho_{coin will be a relation between its total mass M and its total volume V:

\rho_{coin}=\frac{M}{V} (14)

Knowing the total volume of the coin is:

V=V_{gold}+V_{copper}=3.79(10)^{-7}m^{3}+7.6(10)^{-8}m^{3}=4.53(10)^{-7}m^{3} (15)

\rho_{coin}=\frac{0.007988kg}{4.53(10)^{-7}m^{3}} (16)

Finally:

\rho_{coin}=17633.554kg/m^{3}} (17)  This is the total density of the British sovereign coin

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