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NEET · Physics · Units and Dimensions
In the relation E = (α)/(β) e^ -(α z)/(k_B T) , E is energy, z is...
In the relation , is energy, is distance, is the Boltzmann constant and is temperature. The dimensional formula of (where is Planck's constant) is the same as that of :
Options
- AForce
- BAngular momentum
- CPower
- DLinear momentum
Correct answer
D. Linear momentum
Step-by-step solution
According to the principle of homogeneity of dimensions, the exponent must be dimensionless.
\left[\dfrac{\alpha z}{k_B T}\right] = [\text{M}^0\text{L}^0\text{T}^0]k_B T[\text{M}\text{L}^2\text{T}^{-2}][\alpha] = \dfrac{[k_B T]}{[z]} = \dfrac{[\text{M}\text{L}^2\text{T}^{-2}]}{[\text{L}]} = [\text{M}\text{L}\text{T}^{-2}]E\dfrac{\alpha}{\beta}[\beta] = \dfrac{[\alpha]}{[E]} = \dfrac{[\text{M}\text{L}\text{T}^{-2}]}{[\text{M}\text{L}^2\text{T}^{-2}]} = [\text{L}^{-1}]h[\text{M}\text{L}^2\text{T}^{-1}]\beta h[\beta h] = [\text{L}^{-1}] \times [\text{M}\text{L}^2\text{T}^{-1}] = [\text{M}\text{L}\text{T}^{-1}]p = mv \Rightarrow [\text{M}][\text{L}\text{T}^{-1}] = [\text{M}\text{L}\text{T}^{-1}]).
Answer: Linear momentum
\left[\dfrac{\alpha z}{k_B T}\right] = [\text{M}^0\text{L}^0\text{T}^0]k_B T[\text{M}\text{L}^2\text{T}^{-2}][\alpha] = \dfrac{[k_B T]}{[z]} = \dfrac{[\text{M}\text{L}^2\text{T}^{-2}]}{[\text{L}]} = [\text{M}\text{L}\text{T}^{-2}]E\dfrac{\alpha}{\beta}[\beta] = \dfrac{[\alpha]}{[E]} = \dfrac{[\text{M}\text{L}\text{T}^{-2}]}{[\text{M}\text{L}^2\text{T}^{-2}]} = [\text{L}^{-1}]h[\text{M}\text{L}^2\text{T}^{-1}]\beta h[\beta h] = [\text{L}^{-1}] \times [\text{M}\text{L}^2\text{T}^{-1}] = [\text{M}\text{L}\text{T}^{-1}]p = mv \Rightarrow [\text{M}][\text{L}\text{T}^{-1}] = [\text{M}\text{L}\text{T}^{-1}]).
Answer: Linear momentum