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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2010

  • question_answer
    The forbidden energy gap in Ge is 0.72 eV, Given, he = 12400 eV-\[{{A}^{o}}\]. The maximum wavelength of radiation that will generate electron hole pair is

    A) \[172220\overset{\text{o}}{\mathop{\text{A}}}\,\]

    B) \[172.2\overset{\text{o}}{\mathop{\text{A}}}\,\]

    C) \[17222\overset{\text{o}}{\mathop{\text{A}}}\,\]

    D) \[1722\overset{\text{o}}{\mathop{\text{A}}}\,\]

    Correct Answer: C

    Solution :

    Energy gap, \[{{E}_{g}}=\frac{hc}{\lambda }\] \[\lambda =\frac{hc}{{{E}_{g}}}=\frac{12400}{0.72}=17222{\AA}\]


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