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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Chemistry
Chapter
Structure of Atom

JEE Advanced 2019 · Paper 2 · Chemistry · Structure of Atom

The ground state energy of hydrogen atom is - 13.6eV . Consider an...

The ground state energy of hydrogen atom is −13.6eV- 13.6eV . Consider an electronic state ΨΨ of He+{He}^{+} whose energy, azimuthal quantum number and magnetic quantum number are −3.4eV,2- 3.4eV, 2 and 00 respectively. Which of the following statement(s) is(are) true for the state ΨΨ ?

Options

  1. AIt has 22 angular nodes
  2. BIt has 33 radial nodes
  3. CIt is a 4d4d state
  4. DThe nuclear charge experienced by the electron in this state is less than 2e2e , where e is the magnitude of the electronic charge.

Correct answer

A. It has 22 angular nodes

Step-by-step solution

∵EHe+=−13.6(z2)n2=−13.6(2)2n2=−3.4∵ {E}_{H{e}^{+}} = \dfrac{- 13.6\left({z}^{2}\right)}{{n}^{2}} = \dfrac{- 13.6{\left(2\right)}^{2}}{{n}^{2}} = - 3.4
n2=16{n}^{2} = 16
∴n=4∴ n = 4
n=4,l=2,m=0n = 4, l = 2, m = 0 belongs to 4dyz4{d}_{yz} orbital
→\to Radial nodes ⇒n−l−1⇒ n - l - 1
→\to Angular nodes ⇒4−2−1=1⇒ 4 - 2 - 1 = 1
→\to Angular nodes ⇒l⇒2⇒ l ⇒ 2
Since He+H{e}^{+} is a single e−{e}^{-} ion, its nuclear charge experienced.
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