PrepRung
Sign inSign up free

JEE Main 2024 · Chemistry · Atomic Structure · easy

A photon emitted when the electron of Li^2+ drops from n=6 to n=3 has the same energy as the photon emitted in which transition of the hydrogen atom?

Single correct option

A photon emitted when the electron of Li2+\mathrm{Li^{2+}} drops from n=6n=6 to n=3n=3 has the same energy as the photon emitted in which transition of the hydrogen atom?
  1. A. n=2→n=1n=2\to n=1
  2. B. n=3→n=1n=3\to n=1
  3. C. n=3→n=2n=3\to n=2
  4. D. n=4→n=2n=4\to n=2
Show answer and solution

Answer: (A) n=2→n=1n=2\to n=1

Energy ∝Z2(1n12−1n22)\propto Z^2\left(\dfrac1{n_1^2}-\dfrac1{n_2^2}\right). For Li2+\mathrm{Li^{2+}} (Z=3Z=3): 9(19−136)=9×336=0.759\left(\dfrac19-\dfrac1{36}\right)=9\times\dfrac{3}{36}=0.75 (in units of 13.6 eV13.6\ \mathrm{eV}). For hydrogen, 2→12\to1 gives 1−14=0.751-\dfrac14=0.75; 3→13\to1 gives 0.8890.889, 3→23\to2 gives 0.1390.139 and 4→24\to2 gives 0.1880.188. So the match is n=2→n=1n=2\to n=1.
Report a mistake

Original question written in the JEE Main 2024 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake