Haryana PMT Haryana PMT Solved Paper-2011

  • question_answer
    In the radioactive decay law  \[N={{N}_{0}}{{e}^{-}}^{\lambda t}\]the dimensions of \[\lambda \] are

    A)                  \[\left[ {{\text{M}}^{0}}\text{ }{{\text{L}}^{\text{0}}}\text{ }{{\text{T}}^{\text{0}}} \right]~~~~\]                           

    B)  \[\left[ {{\text{M}}^{0}}\text{ }{{\text{L}}^{\text{0}}}\text{ }{{\text{T}}^{-1}} \right]~~~~\]

    C)                  \[\left[ {{\text{M}}^{0}}\text{ }{{\text{L}}^{\text{0}}}\text{ T} \right]~~~~\]                     

    D)  \[\left[ \text{M }{{\text{L}}^{\text{0}}}\text{ }{{\text{T}}^{-1}} \right]~~~~\]

    Correct Answer: B

    Solution :

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


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