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JEE Main & Advanced Physics Photo Electric Effect, X- Rays & Matter Waves Question Bank Critical Thinking

  • question_answer
    When photon of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy TA eV and de-Brolie wavelength \[{{\lambda }_{A}}\]. The maximum kinetic energy of photoelectrons liberated from another metal B by photon of energy 4.70 eV is \[{{T}_{B}}=({{T}_{A}}-1.50)\ eV\]. If the de-Broglie wavelength of these photoelectrons is \[{{\lambda }_{B}}=2{{\lambda }_{A}}\], then                                                              [IIT-JEE 1994]

    A)            The work function of A is 2.25 eV

    B)            The work function of B is 4.20 eV

    C)            \[{{T}_{A}}=2.00\ eV\]     

    D)            \[{{T}_{B}}=2.75\ eV\]

    Correct Answer: B

    Solution :

     Kmax = E ? W0                    \ TA = 4.25 ? (W0)A                                                ...(i)                    TB =(TA? 1.5)= 4.70 ? (W0)B                                                        ...(ii)                    Equation (i) and (ii) gives (W0)B ? (W0)A = 1.95 eV                    De Broglie wave length \[\lambda =\frac{h}{\sqrt{2mK}}\]\[\Rightarrow \lambda \propto \frac{1}{\sqrt{K}}\]                    Þ \[\frac{{{\lambda }_{B}}}{{{\lambda }_{A}}}=\sqrt{\frac{{{K}_{A}}}{{{K}_{B}}}}\] \[\Rightarrow 2=\sqrt{\frac{{{T}_{A}}}{{{T}_{A}}-1.5}}\] Þ TA = 2eV                    From equation (i) and (iii)            WA = 2.25 eV and WB = 4.20 eV.


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