JIPMER Jipmer Medical Solved Paper-1995

  • question_answer
    The wave number corresponding to 430 nm radiation, is:

    A)  \[2.33\times {{10}^{4}}\,c{{m}^{-1}}\]

    B)                         \[2.32\times {{10}^{-4}}\,c{{m}^{-1}}\] 

    C)         \[0.233\times {{10}^{3}}\,c{{m}^{-1}}\]

    D)         \[23.3\times {{10}^{5}}\,c{{m}^{-1}}\]

    Correct Answer: A

    Solution :

    We know that wave number \[(\bar{v})=\frac{1}{\text{wavelength}}\] \[=\frac{1}{430\,nm}=\frac{1}{430\times {{10}^{-7}}cm}\] \[=2.33\times {{10}^{4}}c{{m}^{-1}}\]


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