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JEE Main & Advanced JEE Main Paper (Held On 09-Jan-2019 Morning)

  • question_answer
    Surface of certain metal is first illuminated with light of wavelength \[{{\lambda }_{1}}=350\,nm\] nm and then, by light of wavelength \[{{\lambda }_{2}}=540nm\]. It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to : (Energy of photon \[=\frac{1240}{\lambda \,(in\,\,nm)}\,e\,V)\] [JEE Main Online Paper (Held On 09-Jan-2019 Morning]

    A) 5.6          

    B) 2.5

    C) 1.8                

    D) 1.4

    Correct Answer: C

    Solution :

      \[{{k}_{1}}\,\,=\,\,\frac{1240}{350}\,-\,W\,\,=\,\,3.54\,\,-\,\,W\] \[{{k}_{2}}\,\,=\,\,\frac{1240}{540}-W\,\,\,=\,\,2.3\,\,-\,\,W\] \[\frac{{{k}_{1}}}{{{k}_{2}}}\,=\,\frac{3.5u\,\,-\,\,W}{2.3\,-\,W}\,\,=\,\,\frac{4}{1}\] \[3.54\,\,-W\,\,=\,\,9.2\,\,-\,\,uW\] \[3W\,\,=\,\,5.66\,\,\] \[W\,\,=\,\,1.8\,eV\,\,\] Option (c)


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