NEET Sample Paper NEET Sample Test Paper-50

  • question_answer The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface is 5V. The incident radiation lies in:

    A) X-ray region                  

    B) ultraviolet region

    C) infrared region   

    D) visible region

    Correct Answer: B

    Solution :

    Maximum kinetic energy of electrons \[{{(KE)}_{max.}}=e{{V}_{s}}=5eV\] According to Einstein?s photoelectric equation \[E={{(KE)}_{max.}}\,+\,\,\phi \] where E = Energy of incident radiation. \[\phi =\] Work function \[E=5+6.2=11.2\text{ }eV\] Wavelength of incident radiation, \[\lambda =\frac{hc}{E}\] \[=\frac{1.240\times {{10}^{-6}}}{11.2}\,eV-m\] \[=\frac{1.240\times {{10}^{-6}}}{11.2}\,=0.1107\times {{10}^{-6}}\,m\] \[=1107\,\,\times {{10}^{-10}}m=1107\text{ }\overset{{}^\circ }{\mathop{A}}\,\] The incident radiation lies in the ultraviolet region

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