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JEE Main 202623 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
23 January
Shift
Morning Shift
Subject
Chemistry
Chapter
Structure of Atom

JEE Main 2026 · 23 January · Morning Shift · Chemistry · Structure of Atom

Which of the following statements regarding the energy of the...

Which of the following statements regarding the energy of the stationary state is true in the following one - electron systems ?

Options

  1. A+2.18×10−18J for second orbit of He++2.18 \times 10^{-18} \mathrm{ J} \text{ for second orbit of } \mathrm{He}^{+}ion \text{ion }
  2. B−2.18×10−18J for third orbit of Li2+-2.18 \times 10^{-18} \mathrm{ J} \text{ for third orbit of } \mathrm{Li}^{2+} ion \text{ ion }
  3. C-1.09 \times 10^{-18} \mathrm{ J} \text{ for second orbit of } \mathrm{H} \text{ atom. }
  4. D+8.72×10−18J for first orbit of He++8.72 \times 10^{-18} \mathrm{ J} \text{ for first orbit of } \mathrm{He}^{+}ion \text{ion }

Correct answer

B. −2.18×10−18J for third orbit of Li2+-2.18 \times 10^{-18} \mathrm{ J} \text{ for third orbit of } \mathrm{Li}^{2+} ion \text{ ion }

Step-by-step solution

The energy of an electron in the nthn^{th} stationary state of a one-electron system is given by the Bohr's formula:
En=−2.18×10−18×Z2n2E_n = -2.18 \times 10^{-18} \times \frac{Z^2}{n^2} J
where ZZ is the atomic number and nn is the principal quantum number.
For option (1): Second orbit of He+He^+ (Z=2,n=2Z=2, n=2)
E=−2.18×10−18×2222=−2.18×10−18E = -2.18 \times 10^{-18} \times \frac{2^2}{2^2} = -2.18 \times 10^{-18} J. (Incorrect sign in option)
For option (2): Third orbit of Li2+Li^{2+} (Z=3,n=3Z=3, n=3)
E=−2.18×10−18×3232=−2.18×10−18E = -2.18 \times 10^{-18} \times \frac{3^2}{3^2} = -2.18 \times 10^{-18} J. (Correct)
For option (3): Second orbit of HH atom (Z=1,n=2Z=1, n=2)
E=−2.18×10−18×1222=−0.545×10−18E = -2.18 \times 10^{-18} \times \frac{1^2}{2^2} = -0.545 \times 10^{-18} J. (Incorrect value)
For option (4): First orbit of He+He^+ (Z=2,n=1Z=2, n=1)
E=−2.18×10−18×2212=−8.72×10−18E = -2.18 \times 10^{-18} \times \frac{2^2}{1^2} = -8.72 \times 10^{-18} J. (Incorrect sign in option)
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