2. Solved Papers
  3. JCECE Medical

JCECE Medical JCECE Medical Solved Paper-2013

  • question_answer
    In a photo electric effect experiment, the maximum kinetic energy of the emitted electrons is \[1\,eV\]for incoming radiation of frequency \[{{v}_{0}}\]and \[\text{3 eV}\]for incoming radiation of frequency \[\text{3}{{\text{V}}_{0}}/2.\] What is the maximum kinetic energy of the electrons emitted for incoming radiations of frequency \[\text{9}{{\text{V}}_{0}}/4?\]

    A)  3 eV              

    B)  4.5 eV

    C)  6 eV              

    D)  9 eV

    Correct Answer: C

    Solution :

     \[{{(KE)}_{\max }}=hv-\text{o}{{|}_{0}}\] So, \[1\,eV=h{{v}_{0}}-\text{o}{{|}_{0}}\] ?(i) and \[3\,eV=\frac{h{{v}_{0}}}{2}-\text{o}{{|}_{0}}\] ?(ii) \[\Rightarrow \] \[3eV-1eV=\frac{h{{v}_{0}}}{2}\] or \[h{{v}_{0}}=4\,eV\] From Eq.(i),  \[\text{o}{{|}_{0}}=h{{v}_{0}}-1\,eV\] \[=4eV-1\,eV\] \[=3\,eV\] \[\therefore \] \[{{(KE)}_{\max }}=h\times \frac{9{{v}_{0}}}{4}-3\,eV\] \[=\frac{9}{4}(4\,eV)-3\,eV\] \[=6\,eV\]


You need to login to perform this action.
You will be redirected in 3 sec spinner