Question

In the radioactive decay law $N=N_0{e}^{-\lambda t}$ the dimensions of $\lambda$ are

• A. $[M^0{L}^0{T}^0]$
• B. $[M^0{L}^0{T}^{-1}]$
• C. $[M^0{L}^{0}T]$
• D. $[M{L}^0{T}^{-1}]$

Answer 1

Answer: (B)

$N=N_0{e}^{-\lambda t}$
Exponents are always dimensionless.
So, dimensions of $\lambda t=\left[M^0{L}^0{T}^0\right][\lambda ][T]$
$=\left[M^0{L}^0{T}^0\right][\lambda ]\frac{\left[M^0{L}^0{T}^0\right]}{[T]}$
$=\left[M^0{L}^0{T}^{-1}\right]$