1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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).

You might be interested in
Can someone help me please??
Mrrafil [7]

1. Our solar system is the only place in the universe where gravity played a key part in the formation of planets.

2. Rocky planets are small, dense, and orbit relatively close to the sun, compared to the Jovian planets, which are large, less dense, and orbiting far from the sun.

3. _______

7 0
3 years ago
A wheel accelerates from rest to 34.7 rad/s at a rate of 47.0 rad/s^2. Through what angle (in radians) did the wheel turn while
dem82 [27]

12.8 rad

Explanation:

The angular displacement \theta through which the wheel turned can be determined from the equation below:

\omega^2 = \omega_0^2 + 2\alpha\theta (1)

where

\omega_0 = 0

\omega = 34.7\:\text{rad/s}

\alpha = 47.0\:\text{rad/s}^2

Using these values, we can solve for \theta from Eqn(1) as follows:

2\alpha\theta = \omega^2 - \omega_0^2

or

\theta = \dfrac{\omega^2 - \omega_0^2}{2\alpha}

\:\:\:\:= \dfrac{(34.7\:\text{rad/s})^2 - 0}{2(47.0\:\text{rad/s}^2)}

\:\:\:\:= 12.8\:\text{rad}

7 0
2 years ago
A physicist is investigating a beam of laser light of wavelength 550 nm. The light strikes a target that is 189 meters away from
Lena [83]

Answer: here you go I was looking for this answer everywhere,I have it now so it’s 6.30 x 10^-7 s

Explanation:

I hope this helps☺️

8 0
3 years ago
How society has been benefited by the knowledge of physics​
lilavasa [31]

Answer:

The introduction of machines to make work done more easier and faster

7 0
3 years ago
Read 2 more answers
already did the work, i just need someone to see if i did the tangent line right? the line has to touch the point 0.6! thank you
Dmitrij [34]

The tangent looks good.

The curve is a bit crooked, at the 0.9 and 1.

But overall, cool graph.

5 0
3 years ago
Other questions:
  • If you quadruple the temperature of a black body, by what factor will the total energy radiated per second per square meter incr
    8·1 answer
  • What is the speed of gravity
    15·1 answer
  • I drop a penny from the top of the tower at the front of Fort Collins High School and it takes 1.85 seconds to hit the ground. C
    7·1 answer
  • An airplane on a runway accelerates at 4.0 meters/second2 for 28.0 seconds before takeoff. How far does the plane travel on the
    7·2 answers
  • The part of Earth's rocky outer layer that makes up the land masses is the
    5·1 answer
  • If a car is traveling 60 mph, its tires may have an rpm of 840. How many revolutions do the car’s tires make in one second?
    6·1 answer
  • A 143-g baseball is flying through the air with a speed of 180 km/hr just after it is hit by a bat. If its velocity is at an ang
    7·1 answer
  • An object of mass m moves horizontally, increasing in speed from 0 to v in a time t. The power necessary to accelerate the objec
    9·1 answer
  • 1. What is evaporation and how does it affect weather?
    7·1 answer
  • A test charge gains 10 joules of potential energy as it moves through an electric field. It starts its movement at point 1 and e
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!