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Question: The Richardson equation is given by \(I = A{T^2}{e^{\dfrac{{ - B}}{{kT}}}}\) . The dimensional formu...

The Richardson equation is given by I=AT2eBkTI = A{T^2}{e^{\dfrac{{ - B}}{{kT}}}} . The dimensional formula for AB2A{B^2} is same as that for:
A) IT2I{T^{ - 2}}
B) kTkT
C) Ik2I{k^2}
D) Ik2T\dfrac{{I{k^2}}}{T}

Explanation

Solution

To solve the question, you need to realize that the constant ee is dimensionless and thus, there can be a direct relation between I,AI,A and TT . Similarly, the term in the power of ee is also supposed to be dimensionless since powers cannot have dimensions otherwise the numeric value of the term would mean nothing. Hence, a direct relation between the dimensions of B,kB,k and TT can also be developed rather easily. Once you’ve found out the dimensional relation of AA and BB individually, just find the dimensional formula of the term AB2A{B^2} to reach at the answer.

Complete step by step answer:
We will try to solve the question exactly as explained in the hint section of the solution to the question. For the given equation to be valid, the term as the power of the dimensionless constant ee should also be dimensionless, hence, we will find the dimensional formula of BB in terms of kk and TT from there. As for the dimensional formula of AA , we have already seen that the constant ee is dimensionless, hence, there is a direct relation between the dimensions of AA with the dimensions of II and TT .
Let us find the dimensions of AA first:
Let us talk about the dimensions of the given equation. For it to be valid, the dimensions on both the sides of the equation should be exactly the same. We can see that ee is dimensionless, hence, the only terms with dimensions can be written as:
I=AT2I = A{T^2}
From this equation, we can easily find the dimensions of AA in terms of II and TT as:
A=IT2A = \dfrac{I}{{{T^2}}}
Now, let us find the dimensions of BB :
We have already talked about the fact that the powers are always supposed to be dimensionless, hence, if we talk about the term as the power to the constant ee , we can safely say that this term should be dimensionless, hence, we can write the equation, in terms of dimensions as:
BkT=1\dfrac{B}{{kT}} = 1
So, we can write the dimensions of BB in terms of kk and TT as:
B=kTB = kT
Now, the question has asked us about the dimensional formula of AB2A{B^2} . To find this, we simply have to substitute the dimensional relations of AA and BB as:
AB2=(IT2)(kT)2A{B^2} = \left( {\dfrac{I}{{{T^2}}}} \right){\left( {kT} \right)^2}
Upon solving it, we will get:
AB2=Ik2A{B^2} = I{k^2}

Hence, we can see that the option (C) is the correct option as the value matches with the one that we found out.

Note: Many students try to manipulate the given equation in hopes of finding a direct relation between AB2A{B^2} and the remaining other terms, which would never work in such questions. Also remember that in such questions, any symbol’s dimensions cannot be presumed to be something even if you know the symbols, this would only confuse you and make you lose marks.