Contents
Some of the most important Physics Topics include energy, motion, and force.
What is the Energy Conversion of Nuclear Energy?
Introduction
Physicist Ernest Rutherford was able to reach two important conclusions from his famous alpha particle scattering experi-ment, regarding distribution of charge carriers and mass in an atom:
i) The entire positive charge and most of the mass of an atom is concentrated in a very small space of the atom called the nucleus. The volume of the nucleus is only about 1 in 1012 part of the atom (atomic diameter is about 10-8 cm and diameter of nucleus is estimated as 10-12 cm).
ii) The remaining part of the atom contains negatively charged electrons. These electrons are distributed in a regular pat-tern outside the nucleus; and a large part of the atom is empty space. Total mass of the electrons is negligible in comparison to the mass of the atom.
Mass-Energy Equivalence
Matter can be viewed as concentrated energy. Max Planck and others had realised the importance of the concept, early in twen-tieth century but it was Albert Einstein who first proposed an equivalence of mass and energy. He suggested c2 (c = velocity of light) as the conversion factor from mass to energy.
The principle of mass-energy equivalence can be stated thus : If the mass m of a body is completely converted to energy, the amount of the energy is
E = mc2 [c = speed of light in vacuum = constant]
The relation suggests that the energy E is equivalent to mass m or that the mass m is equivalent to energy E.
Example:
i) In CGS system,
c = 2.998 × 1010cm ᐧ s-1 \(\simeq\) 3 × 1010 cm ᐧ s-1
∴ Equivalent energy contained in 1 g,
E = 1 × (3 × 1010)2 = 9 × 1020 erg = 9 × 1013 J
Again in SI, c = 3 × 108m ᐧ s-1
∴ Equivalent energy contained in 1 kg,
E = 1 kg × (3 × 108m ᐧ s-1)2 = 9 × 1016J
ii) Mass of electron, me = 9.109 × 10-28g
∴ Equivalent energy of mass of an electron
= 9.109 × 1o-28 × (3 × 1010)2 erg
= \(\frac{9.109 \times 10^{-28} \times 9 \times 10^{20}}{1.602 \times 10^{-12}}\)eV
= 0.511 × 106 eV = 0.511 MeV
Rest mass: Einstein’s theory of relativity also-suggests that mass of a body is not a constant but depends on the velocity of the body. Specially when speed is comparable to the speed of light in vacuum, mass of a body increases considerably. Hence, when mentioning the mass as an innate property of matter, the body should be considered to be at rest. This is called rest mass. For example rest mass of an electron = 9.109 × 10-28g. Lorentz derived the relation between rest mass (m0) and the mass at velocity v close to c as
m = \(\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}\)
Unit of mass and energy: As mass and energy are equivalent to each other, their units too are equivalent. Hence, energy unit is also used to represent mass and vice versa. For example, 1 g energy denotes 9 × 1020 erg of energy or a mass of 9 × 1016 J indicates 1 kg mass. Using this equivalence of mass and energy we can say, rest mass of an electron, me = 0.511 MeV.
Law of conservation of mass-energy: when there occurs an interconversion between mass and energy, the law of conservation of mass and the law of conservation of energy can-not be applied separately. Instead, these laws combine to form the law of conservation of mass-energy.
In nature, the sum of mass and energy of a system is a con-stant. While there may be various changes in the form, mass- energy cannot be destroyed or created.
Atomic energy: Conversion of mass into energy can take place only in nuclear phenomenon within atoms. Energy from this conversion is the source of atomic energy. Atomic energy is used in making nuclear weapons like atom bomb, hydrogen bomb, in generating electricity in nuclear power stations etc.
Numerical Examples
Example 1.
In any nuclear reaction \(\frac{1}{1000}\) part of the mass of a particular substance Is converted into energy. If 1 g of that substance takes part In a nuclear reaction then determine the energy evolved in kilowatt-hour.
Solution:
Energy converted = \(\frac{1}{1000}\) × 1 = 0.001 g
∴ Energy evolved,
E = mc2 = 0.001 × (3 × 1010)2
= 9 × 1017 erg = 9 × 1010J = 9 × 1010 W ᐧ s
= \(\frac{9 \times 10^{10}}{1000 \times 3600}\) kW ᐧ h = 25000 kW ᐧ h
Example 2.
If a metal’s mass could be completely convened into energy, calculate how much of this metal would be required as fuel for a power plant in a year. The power plant, let us suppose, generates 200 MW on an average.
Solution:
200 MW = 200 × 106W = 2 × 108W,
1 year = 365 × 24 × 60 × 60 s
∴ In 1 year the energy generated,
E = 2 × 108 × 365 × 24 × 60 × 60 J
∴ Equivalent mass,
m = \(\frac{E}{c^2}\) = \(\frac{\dot{2} \times 10^8 \times 365 \times 24 \times 60 \times 60}{\left(3 \times 10^8\right)^2}\)
= 0.070 kg = 70 g
Principle of mass-energy equivalence can be obtained from Einstein’s famous theory of relativity.
Mass is a form of energy. Mass can be converted into energy and vice versa. If m amount of mass of a substance is completely converted to energy, then the amount of energy liberated is, E = mc2 (c = velocity of light in vacuum = a constant)
According to this equation, equivalent energy of mass m is E, and equivalent mass of energy E is m.
Example:
In CGS system c = 3 × 1010 cm ᐧ s-1
∴ Equivalent energy of 1 g mass
= 1 × (3 × 1010)2 = 9 × 1020 erg = 9 × 1031J.
Again in SI, c = 3 × 108 m ᐧ s-1
Hence equivalent energy of 1 kg mass
= 1 × (3 × 108)2 = 9 × 1016J.
ii) Mass of an electron = 9.1 × 10-28 g
Equivalent energy of the mass of 1 electron
= 9.1 × 10-28 × (3 × 1010)2 erg
= \(\frac{9.1 \times 10^{-28} \times 9 \times 10^{20}}{1.6 \times 10^{-12}}\)erg
= 0.511 × 106 eV = 0.511 MeV.
[The increase or decrease in energy of an electron while crossing a potential difference of 1 V is called 1 electronvolt (eV). 1eV = 1.6 × 10-12 erg.
Rest mass: Einstein’s theory of relativity also informs us that mass of a substance is not a constant quantity but depends on the velocity of the substance. Specially if the velocity of an object becomes comparable to that of light, then its mass increases significantly. That is why to measure the true mass of an object as an intrinsic property, the object should be at rest with respect to the observer. The mass of the object thus measured is called its rest mass and the equivalent energy is called rest energy For example, rest mass of an electron = 9.1 × 10-28 g and rest energy is 0.511 MeV.
Unit of mass and energy: As mass and energy are equivalent to each other, their units are also the same. Some times mass is given in units of energy and energy is given in units of mass. For example, ‘energy of 1 g’ denotes 9 × 1020 erg amount of energy; or mass of ‘9 × 1016 J’ represents a mass of 1 kg. From this equivalence it can be stated that the rest mass of an electron is 0.511 MeV.
Law of conservation of mass-energy: During the mutual transformation of mass and energy, the law of conservation of mass or the law of conservation of energy cannot be applied separately. It becomes the law of conservation of mass-energy.
The total amount of mass-energy in nature is constant, it can never be created or destroyed. It can only change from one form to another.
Mass can be transformed to energy only within an atom. Energy thus obtained from mass is the source of atomic energy. When gamma rays with an energy of a few MeV or more enters the electric field of a heavy nucleus, it can be transformed into an electron and a positron (an electron-like particle but of positive charge). This is an example of transformation of energy to mass.
Rotational Motion of a Body in a Vertical Circle
When a body of mass m is tied to a string and is rotated in a vertical circle, it will not rotate with uniform speed in that circular path [Fig.]. So, it is an example of a non-uniform circular motion.
The magnitude of its velocity goes on increasing when it descends from the highest point B of the circle to the lowest point A. Again, the magnitude of velocity gradually decreases when it rises up to the highest point from the lowest .
During its rotation, the weight of the body always acts vertically downwards, but the direction of centripetal force changes continuously. Hence, the tension in the string does not remain constant.
At the highest point B of the circular path, tension in the string and the weight of the body together provide the necessary centripetal force for the rotation of the body. If the velocity of the body is u and the tension in the string is T2 at the point B, then
T2 + mg = \(\frac{m v^2}{r}\)
or, T2 = \(\frac{m v^2}{r}\) – mg ……… (1)
Again, the tension in the string at the lowest point A on the circular path acts in the vertically upward direction, i.e., towards the centre of the circle, but the weight of the body acts vertically downwards. As a result, the difference between the tension in the string and the weight of the body provides the necessary centripetal force. If the velocity of the body is u and the tension in the string is T1 at the point A, then
T1 – mg = \(\frac{m u^2}{r}\)
or, T1 = \(\frac{m u^2}{r}\) + mg ……… (2)
If the tension in the string at the point B is zero, i.e., if T2 = 0, then
\(\frac{m v^2}{r}\) – mg = 0 or, \(\frac{m v^2}{r}\) = mg
or, v = \(\sqrt{r g}\) …….. (3)
Hence, if the value of v is less than \(\sqrt{r g}\) at the highest point, then the tension in the string becomes negative, i.e., the string gets relaxed. As a result, the body descends in a parabolic path instead of rotating along a circular path.
Hence, the least velocity of the body at the highest point on the circular path to maintain the vertical circular motion should be \(\sqrt{r g}\). This least velocity is called the Critical velocity.
Least velocity at the lowest point on the circular path necessary to maintain the critical velocity: If the potential energies of the body at points A and B are V1 and V2, respectively, then
total energy at point A = \(\frac{1}{2} m u^2\) + V1.
Similarly, total energy at point B = \(\frac{1}{2} m u^2\) + V2.
According to the principle of conservation of energy,
\(\frac{1}{2} m u^2\) + V1 = \(\frac{1}{2} m v^2\) + V2
or, V2 – V1 = \(\frac{1}{2}\)m(u2 – v2) ………. (4)
When a body of mass m attains a height h, the potential energy gained by it = mgh. In the given figure, the height of the point B with respect to the point A is 2r.
So, the change in potential energy = V2 – V1 = mg ᐧ 2r
∴ From equation (4) we get,
mg ᐧ 2r = \(\frac{1}{2}\)m(u2 – v2)
or, 4gr = u2 – v2
or, u2 = 4gr + v2 = 4gr + gr [∵ the least velocity at B, v = \(\sqrt{g r}\)]
or, u2 = 5gr
or, u = \(\sqrt{5 g r}\) ….. (5)
For rotation in a vertical circle, this is the least value of u. It should be noted that the value of this least velocity is independent of the mass of the body.
For this least value of u, the minimum value of tension at the point A can be determined from the equation (2).
(T1)min = \(m u_{\min }^2\) + mg = \(\frac{m \cdot 5 g r}{r}\) + mg
= 5mg + mg = 6mg
or, (T1)min = 6 × weight of the body ….. (6)
So, if a body is rotated in a vertical circle, then at the lowest point of the circle, the minimum tension in the string is 6 times the weight of the body.
Rotation of a bucket M of water in a vertical circle: When a bucket full of water is made to rotate with high speed in a vertical circle, it is seen that at the highest point of the circular path, though the bucket is inverted, water from it does not spill out [Fig.]. It is clear that when the bucket is at the highest position, two forces act simultaneously on the water in the bucket:
i) Downward weight (mg) of water (m = mass of water).
ii) Upward reaction of the centripetal force; if the velocity of the bucket at the highest position is v and the radius of the circular path is r, the value of this centripetal force acting on the bucket due to its rotation = \(\frac{m v^2}{r}\)
Thus, when the upward force becomes greater than the force downward, water in the bucket cannot fall down. So, the condition for water to not fall down is,
\(\frac{m v^2}{r}\) ≥ mg or, v2 ≥ rg
Hence, the minimum value of v should be \(\sqrt{r g}\).
According to equation (6), when this bucket is rotated along a complete circular path, at the lowest position of the bucket, the tension acting on the hand will be 6 times the weight of the bucket filled completely with water. Hence, it is difficult to perform the experiment with a heavy bucket full of water. However, it is easier with a bucket or some other container containing a little amount of water.
Numerical Examples
Example 1.
A body of mass 1 kg is tied with a thread and is whirled in a vertical circle of radius 50 cm with a speed of 500 cm ᐧ s-1. What will be the tension in the thread at the highest and the lowest positions of the body?
Solution:
Speed of the body at every point on the circular path is the same and its value, v = 500 cm ᐧ s-1 = 5 m ᐧ s-1
So, at every point on the circular path,
centripetal force, F = \(\frac{m v^2}{r}\) = \(\frac{1 \times(5)^2}{0.5}\) = 50 N
Weight of the body = mg = 1 × 9.8 = 9.8 N
At the highest position, the tension in the thread (T1) and weight of the body both act downwards and together they provide the necessary centripetal force. Hence,
F = mg + T1 or, T1 = F – mg = 50 – 9.8 = 40.2N
At the lowest position, the tension in the thread (T2) acts in the upward direction but the weight acts downwards, and hence, in this case, (T2 – mg) provides the necessary centripetal force.
So, T2 – mg = F or, T2 = F + mg = 50 + 9.8 = 59.8 N.
Example 2.
A particle of mass 100 g is suspended from the end of a weightless string of length 100 cm and is rotated in a vertical plane. The speed of the particle is 200 cm ᐧ s-1 when the string makes an angle of θ = 60° with the vertical. Determine
(i) the tension in the string when ú = 60° and
(ii) the speed of the particle at the lowest position. Acceleration due to gravity = 980 cm ᐧ s-2.
Solution:
i) The body is rotated in a vertical plane [Fig.].
At point B, when the string makes an angle 60° with the vertical, the velocity of the particle is v (say).
If the tension in the string is T, then
T – mgcos6o° = \(\frac{m v^2}{r}\)
∴ T = mgcos60° + \(\frac{m v^2}{r}\)
Here, v = 200 cm ᐧ s-1; r = 100 cm; m = 100 g
∴ T = 100 × 980 × \(\frac{1}{2}\) + \(\frac{100 \times(200)^2}{100}\)
= 8.9 × 104 dyn.
ii) From the principle of conservation of energy we get, kinetic energy of the body at the lowest point A = sum of the potential and kinetic energies at the point B
or, \(\frac{1}{2} m v_A^2\) = \(\frac{1}{2} m v^2\) + mg ᐧ AC
[vA = speed of the particle at the point A]
or, \(v_A^2\) = v2 + 2g ᐧ AC
Here, AC = OA – OC = OA – OBcosθ
= 1 – icosθ = 1oo – 100cos60° = 50 cm
∴ \(v_A^2\) = (200)2 + 2 × 980 × 50 = 13.8 × 104
or, vA = 371.5 cm ᐧ s-1.
Example 3.
A body slides down an inclined plane after being released from rest from a height h and finally It
describes a circle of radius r instead of travelling along a horizontal floor. For what minimum value of h, can the particle describe that motion? Ignore friction.
Solution:
After descending along the inclined plane, the body can describe the circular path if the velocity of the body at the lowest point B of the circular path is, at least, C
v = \(\sqrt{5 g r}\) [Fig.]
Kinetic energy of the body at the point B
= \(\frac{1}{2} m v^2\) = \(\frac{1}{2} m \cdot 5 g r\) = \(\frac{5}{2} m g r\)
The falling body will acquire this kinetic energy in exchange of the potential energy that it possesses at the height h.
So, \(\frac{5}{2} m g r\) = mgh or, h = \(\frac{5}{2} r\)
Example 4.
A body starts falling from the top of a smooth sphere of radius r. What angle does the body subtend at the centre of the sphere when it just loses contact with the sphere and what will be Its velocity then?
Solution:
Let the angle subtended by the body at the centre of the sphere when it loses contact with the sphere be θ.
The body just loses contact with the sphere when the component of gravitational force cannot provide the necessary centripetal force to the body so that it can continue in the circular path.
So, according to Fig.,
mgcosθ = \(\frac{m v^2}{r}\) ….. (1)
(v = velocity of the body when it just loses contact with the sphere)
According to the principle of conservation of energy,