Pressure P varies as P=αβ exp(−αKBθZ) where Z denotes distance,KB Boltzman's constant,θ absolute temperature and α,β are constants.Derive the dimensions of β
[M0L0T0]
[M0L2T0]
[M−1LT−2]
[M0L−1T−2]
A
[M0L0T0]
B
[M0L2T0]
C
[M−1LT−2]
D
[M0L−1T−2]
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Solution
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P=αβ exp(−αkBθZ)
Since the terms inside the exponential do not have any dimension,we can find out the dimension of α
α×L′×kJ×k
α has dimension of JL′=ML2T−2L′=MLT−2
For αβ to have dimension of pressure ,i.e ML−1T−2
β should have the dimension L2 i.e,M0L2T0
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