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JEE Main 2016 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

The photon emitted when the electron of a doubly ionised lithium ion Li^2+ jumps from the n=3 to the n=2 level falls on a metal of work function 4…

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The photon emitted when the electron of a doubly ionised lithium ion Li2+\mathrm{Li^{2+}} jumps from the n=3n=3 to the n=2n=2 level falls on a metal of work function 4 eV4\ \mathrm{eV}. Using the hydrogen ground state energy −13.6 eV-13.6\ \mathrm{eV}, the maximum kinetic energy of the photoelectrons is
  1. A. 9.0 eV9.0\ \mathrm{eV}
  2. B. 13.0 eV13.0\ \mathrm{eV}
  3. C. 17.0 eV17.0\ \mathrm{eV}
  4. D. 21.0 eV21.0\ \mathrm{eV}
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Answer: (B) 13.0 eV13.0\ \mathrm{eV}

For a hydrogen-like ion En=−13.6Z2n2 eVE_n=-13.6\dfrac{Z^2}{n^2}\ \mathrm{eV}. With Z=3Z=3, the photon energy is hν=13.6×9(14−19)=13.6×9×536=17.0 eVh\nu=13.6\times9\left(\dfrac14-\dfrac19\right)=13.6\times9\times\dfrac{5}{36}=17.0\ \mathrm{eV}. By the photoelectric equation, Kmax=hν−ϕ=17.0−4=13.0 eVK_{max}=h\nu-\phi=17.0-4=13.0\ \mathrm{eV}. Quoting the photon energy as the kinetic energy ignores the work function, while adding ϕ\phi instead of subtracting it gives 21 eV21\ \mathrm{eV}.
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