Question

If P represents radiation pressure, C represents the speed of light, and Q represents radiation energy striking a unit area per second, the non-zero integers x, y and z such that ${ P }^{ x }.{ Q }^{ y }.{ C }^{ z }$ is dimensionless, find the values of x, y and z.

Answer 1

Hint: Write down the dimensions of radiation pressure, speed of light and radiation energy. Substitute them in the given equation which is dimensionless. Apply the principle of homogeneity and get the equations. Solve those equations and find the value of x, y and z.

Complete step by step answer:
Given: ${ \left[ P \right] }^{ x }{ \left[ Q \right] }^{ y }{ \left[ C \right] }^{ z }$ is dimensionless
$\therefore { \left[ P \right] }^{ x }{ \left[ Q \right] }^{ y }{ \left[ C \right] }^{ z }=\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$ …(1)

Dimensions of Radiation pressure are $\left[ P \right] =\left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right]$ …(2)
Dimensions of Speed of light is $\left[ C \right] =\left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 } \right]$ …(3)
Dimensions of Radiation Energy is $\left[ Q \right] =\left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -3 } \right]$ …(4)

Substituting equation. (2), equation. (3) and equation. (4) in equation. (1) we get,
${ \left[ { M }^{ 1 }{ L }^{ -1 }{ T }^{ -2 } \right] }^{ x }{ \quad \left[ { M }^{ 1 }{ L }^{ 0 }{ T }^{ -3 } \right] }^{ y }\quad { \left[ { M }^{ 0 }{ L }^{ 1 }{ T }^{ -1 } \right] }^{ z }=\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$
$ \Rightarrow { \left[ { M }^{ x }{ L }^{ -x }{ T }^{ -2x } \right] }{ \quad \left[ { M }^{ y }{ L }^{ 0 }{ T }^{ -3y } \right] }\quad { \left[ { M }^{ 0 }{ L }^{ z }{ T }^{ -z } \right] }=\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$
$ \Rightarrow { \left[ { M }^{ x+y }{ L }^{ -x+z }{ T }^{ -2x-3y-z } \right] }=\quad \left[ { M }^{ 0 }{ L }^{ 0 }{ T }^{ 0 } \right]$

Now, by equating the powers on the left side with the right side,
$ \quad x\quad +\quad y=\quad 0 $…(1)
$ \quad -x\quad +\quad z=\quad 0$ …(2)
$ \quad -2x\quad -\quad 3y\quad -\quad z=\quad 0$ …(3)

Solving equation. (1), (2) and (3) we get,
x= 1, y= -1, z= 1
Hence, the values of x, y and z are 1, -1 and 1 respectively.

Note:
Take care while writing dimensions of radiation energy. You can’t write dimensions the same as that of energy. Dimensions of energy and radiation energy are not the same. As it is mentioned in the question, it is radiation energy per unit area and per sec.