JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    The energy in \[eV\] of red light of wavelength \[\lambda =6560\overset{\text{o}}{\mathop{\text{A}}}\,\] is:

    A) \[1.89\,\,eV\]                   

    B) \[2.89\,\,eV\]

    C) \[3.89\,\,eV\]                   

    D) \[4.89\,\,eV\]

    Correct Answer: A

    Solution :

    From Planck's law, the energy \[(E)\] of a wave of wavelength \[\lambda \] is                 \[E=\frac{hc}{\lambda }\] where, \[h\] is Planck's constant, \[c\] is speed of light, and \[\lambda \] is wavelength. Given,   \[h=6.625\times {{10}^{-34}}J\text{-}s\] \[c=3\times {{10}^{8}}m/s,\,\,\lambda =6560{\AA}=6560\times {{10}^{-10}}m\],                 \[1\,\,eV=1.6\times {{10}^{-19}}J\] \[\therefore \]  \[E=\frac{6.625\times {{10}^{-34}}\times 3\times {{10}^{8}}}{6560\times {{10}^{-10}}\times 1.6\times {{10}^{-19}}}eV\] \[\Rightarrow \]               \[E=1.89\,\,eV\]


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