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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    A photon of wavelength 300 nm interacts with a stationary hydrogen atom in ground state. During the interaction, whole energy of the photon is transferred to the electron of the atom. State which possibility is correct. (Consider, Plank constant \[=4\times {{10}^{-15}}\,eVs,\] velocity of light \[=3\times {{10}^{8}}m/s,\] ionization energy of hydrogen\[=13.6\text{ }eV\])

    A)  Electron will be knocked out of the atom

    B)  Electron will go to any excited state of the atom

    C)  Electron will go only to first excited state of the atom

    D)  Electron will keep orbiting in the ground state of the atom

    Correct Answer: D

    Solution :

     The energy of the photon \[E=\frac{hc}{\lambda }\] \[=\frac{4\times {{10}^{-15}}eVs\times 3\times {{10}^{8}}m/s}{300\times {{10}^{-9}}m}\] \[=\frac{4\times {{10}^{2}}}{300}eV=\frac{4}{3}eV=1.33\,eV\] The ionisation energy is 13.6 eV which is greater than energy of photon, so atom can not come into excited state and will remain in ground state.


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