Question

The wavelength associated with a moving particle depends upon the power \[p\]of its mass\[m\],\[{q^{th}}\] power of its velocity \[v\] , and \[{r^{th}}\] power of Planck's constant \[h\]. Then the correct set of values of\[p,q\;\] and \[r\;\] is:-
A. \[p = 1,q = - 1,r = - 1\]
B. \[\;p = 1,q = 1,r = 1\]
C. \[\;p = - 1,q = - 1,r = - 1\]
D. \[\;p = - 1,q = - 1,r = 1\]

Answer 1

Hint: Dimensional analysis is defined as the study of the relationship between the physical quantities with the help of the dimensions and units. The principle of homogeneity is the principle used in dimensional analysis. It states that both sides of the dimensional equation should contain terms that have the same dimensions. This principle is very useful because it can help us to convert the units from one form to another.


Complete step by step solution:
Given that the wavelength associated with the moving particle depends upon its mass, velocity and Planck's constant. We can write this mathematically as,
\[\lambda \propto mvh\]
The wavelength is associated with some power of each term in the R.H.S
So we can write that as,
\[\lambda \propto {m^p}{v^q}{h^r}\]
Removing the proportionality constant and replacing it with a term. Let the term be
\[\lambda = K{m^p}{v^q}{h^r}\]
Now we can write the dimension of each term.
Dimension of wavelength, \[\lambda = \left[ L \right]\]
Dimension of mass, \[m = \left[ M \right]\]
Dimension of velocity,\[v = \left[ {L{T^{ - 1}}} \right]\]
Dimension of Planck constant, \[h = \left[ {M{L^2}{T^{ - 1}}} \right]\]
Now applying the homogeneity principle and equating the dimensions of both the sides of the equations we get,
\[[L] = K{[M]^p}{[L{T^{ - 1}}]^q}{[M{L^2}{T^{ - 1}}]^r}\]
\[[L] = K{[M]^{(p + r)}}{[L]^{(q + 2r)}}{[T]^{(q - r)}}\]
Comparing the powers on both sides,
\[p + r = 0\]
\[q + 2r = 1\]
\[ - q - r = 0\]
Thus we get
\[p = - 1;q = - 1;r = 1\]
Therefore the correct option is D.

Note:
Dimensional analysis can also be called the Factor Label Method or Unit Factor Method. Because we use these conversion factors to get the same units. Let us understand this using an example. Let us convert \[5{\text{ }}km\]to \[m\]. We know that \[1000{\text{ }}meters\] is equal \[1km\]. Therefore,
\[5km = 5 \times 1000m = 5000meters\]. Hence here the conversion factor is \[1000{\text{ }}meters\]