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JEE Main & Advanced JEE Main Paper (Held On 10-Jan-2019 Evening)

  • question_answer
    A metal plate of area \[1\times {{10}^{-}}^{4}\,{{m}^{2}}\] is illuminated by a radiation of intensity\[16\text{ }mW/{{m}^{2}}\]. The work function of the metal is 5 eV. The energy of the incident photons is 10 eV and only \[10%\] of it produces photo electrons. The number of  emitted photoelectrons per second and their maximum energy, respectively, will be- [JEE Main Online Paper (Held On 10-Jan-2019 Evening]

    A) \[{{10}^{14}}\] and 10 eV                   

    B) \[{{10}^{12}}\] and 5 Ev

    C) \[{{10}^{11}}\] and 5 eV

    D)                  \[{{10}^{10}}\]and 5 eV

    Correct Answer: D

    Solution :

    Energy incident on plate per second \[=\,\,IA\] \[~=16\times 10-3\times 1\times {{10}^{-}}^{4}\] \[=\,\,16\times {{10}^{-}}^{7}watt\] \[K{{E}_{max}}=hf-\phi \] \[=\text{ }10-5=5\,eV\] \[\frac{\overset{\bullet }{\mathop{N}}\,\,hc}{\lambda }=1.6\times {{10}^{-\,7}}\] \[\overset{\bullet }{\mathop{N}}\,\times 10\times 1.6\times {{10}^{-19}}\,=1.6\times {{10}^{-7}}\] \[\overset{\bullet }{\mathop{N}}\,\times \text{1}{{\text{0}}^{-18}}\,={{10}^{-7}}\] \[\overset{\bullet }{\mathop{N}}\,={{10}^{11}}\] \[10%\] of incident photon emit electron \[\therefore \] No of emitted electrons per second \[=\,\,\overset{\bullet }{\mathop{N}}\,\,\times \,\frac{10}{100}\,\,=\,\,{{10}^{10}}\]


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