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NEET Sample Paper NEET Sample Test Paper-6

  • question_answer
    The relationship between the energy of the radiation with a wavelength \[8000\overset{o}{\mathop{A}}\,\] and the energy \[{{E}_{2}}\] of the radiation with a wavelength \[16000\text{ }\overset{o}{\mathop{A}}\,\] is:

    A)  \[{{E}_{1}}={{E}_{2}}\]              

    B)  \[{{E}_{1}}=4{{E}_{2}}\]

    C)  \[{{E}_{1}}=\frac{1}{2}{{E}_{2}}\]                   

    D)  \[{{E}_{1}}=2{{E}_{2}}\]

    Correct Answer: D

    Solution :

    \[\because \]     \[E=\frac{hc}{\lambda }\] \[\therefore \]  \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{16000}{8000}\]                 \[\frac{{{E}_{1}}}{{{E}_{2}}}=\frac{2}{1}\]                 \[{{E}_{1}}=2{{E}_{2}}\]


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