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AMU Medical AMU Solved Paper-2015

  • question_answer
    Ultraviolet light of wavelength 350 nm and intensity \[1.00\text{ }W/{{m}^{2}}\]is directed at a potassium surface. What will be the maximum kinetic   energy of the photoelectrons? (Work function of potassium =2.2eV)

    A)  3.5eV                                  

    B)  2.6eV

    C)  2.2eV                                  

    D)  1.3eV

    Correct Answer: D

    Solution :

                     Given, \[\lambda =350\,nm=350\times {{10}^{-7}}m\] Intensity \[=1.00\,\,\omega /{{m}^{2}}\] work function W = 2.2 eV Energy of the photon                 \[=\frac{hc}{\lambda }\]                 \[=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{350\times {{10}^{8}}}\]                 \[=3.5\,eV\]                 \[{{K}_{\max }}=E-W\]                 \[=(3.5-2.2)\,eV\]                 \[=1.3\,eV\]


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