Question

1 mg of matter into energy will give
(A) 90 joules
(B) $9 \times {10^3}$ Joules
(C) $9 \times {10^5}$ Joules
(D) $9 \times {10^5}$ Joules

Answer 1

The energy in any given matter can be defined as the product of its mass and the square of the speed of light. All quantities should be used in SI units.

Formula used: In this solution we will be using the following formulae;
$E = m{c^2}$ where $E$ is the energy in a given mass, $m$ is the mass of the substance, and $c$ is the speed of light.

Complete Step-by-Step solution:
Generally, all objects possess energy even at its rest position, and away from any field.
In the question above, we are to find the energy of the body due to its mass alone. This energy is simply the product of the mass of the body and the square of the speed of light. This is mathematically given as
$E = m{c^2}$ where $E$ is the energy in a given mass, $m$ is the mass of the substance, and $c$ is the speed of light.
The mass must first be converted to SI unit, hence
$m = 1mg = 0.000001kg = 1.0 \times {10^{ - 6}}kg$
Hence, the energy is
$E = 1.0 \times {10^{ - 6}} \times {\left( {3.0 \times {{10}^8}} \right)^2} = {10^{ - 6}} \times 9 \times {10^{16}}$
$E = 9 \times {10^{10}}J$
Hence, the correct option is D.

Note:
For clarity, the above matter is assumed to be constant. When the object is moving, it additionally possesses a kinetic energy in which when the matter is converted to energy is added to the rest energy of the matter (the rest energy is what we just calculated above). For non-relativistic movement, we calculate the total energy to be
$E = m{c^2} + \dfrac{1}{2}m{v^2}$ where $v$ is simply the velocity, and the whole $\dfrac{1}{2}m{v^2}$ is the kinetic energy of the body. For relativistic movement, the total energy is more complicated.