PrepRung

JEE Advanced 2018 Paper 1 · Physics · Atoms, Nuclei and Dual Nature of Radiation · medium

A hydrogen-like ion Li^2+ (Z=3) in its second excited state makes a transition to the first excited state. The emitted photon falls on a metal plate…

One or more options correct

A hydrogen-like ion Li2+\mathrm{Li^{2+}} (Z=3Z=3) in its second excited state makes a transition to the first excited state. The emitted photon falls on a metal plate of work function 14.0 eV14.0\ \mathrm{eV}. Use En=−13.6Z2/n2 eVE_n=-13.6Z^2/n^2\ \mathrm{eV} and hc=1240 eV nmhc=1240\ \mathrm{eV\,nm}. Which of the following is/are correct?
  1. A. The photon energy is 1.89 eV1.89\ \mathrm{eV}.
  2. B. The maximum kinetic energy of the emitted photoelectrons is 3.0 eV3.0\ \mathrm{eV}.
  3. C. The stopping potential is 17 V17\ \mathrm{V}.
  4. D. The wavelength of the photon is about 730 nm730\ \mathrm{nm}.
Show answer and solution

Answer: B

The second excited state is n=3n=3 and the first excited state is n=2n=2. The photon energy is E=13.6×9(14−19)=17.0 eVE=13.6\times9\left(\tfrac14-\tfrac19\right)=17.0\ \mathrm{eV}; the value 1.89 eV1.89\ \mathrm{eV} is the hydrogen (Z=1Z=1) Balmer value of the same transition, so (A) is false. The photoelectric equation gives Kmax=17.0−14.0=3.0 eVK_{max}=17.0-14.0=3.0\ \mathrm{eV}, so (B) is true. The stopping potential is Kmax/e=3.0 VK_{max}/e=3.0\ \mathrm{V}, not 17 V17\ \mathrm{V}, so (C) is false. The wavelength is λ=1240/17.0≈73 nm\lambda=1240/17.0\approx73\ \mathrm{nm}, which is in the ultraviolet and not 730 nm730\ \mathrm{nm}, so (D) is false.
Report a mistake

Original question written in the JEE Advanced 2018 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake