ABC26PH0270 · Atomic Structure

Subject: Physical Chemistry · Chapter: Atomic Structure · Topic: Atomic Structure – General · Exam: GATE 2005-2021 · Marks: 2 · Difficulty: Medium

To demonstrate the variational principle, a trial function $\psi=\mathrm{C}_{1} \frac{\psi_{2 \mathrm{s}}+\psi_{3 \mathrm{s}}}{\sqrt{2}}+\mathrm{C}_{2} \frac{\psi_{2 \mathrm{s}}-\psi_{3 \mathrm{s}}}{\sqrt{2}}$ where $\mathrm{C}_{1}$ and $\mathrm{C}_{2}$ are the variational parameters and $\Psi_{2 \mathrm{s}}$ and $\Psi_{3 \mathrm{s}}$ are the 2s, and 3s orbitals of the hydrogen atom, is constructed. The corresponding secular determinant for the hydrogen atom (in eV) is
(a)$\left|\begin{array}{cc} 3.4(1+4/9)/2 - E & 3.4(1-4/9)/2 \\ 3.4(1-4/9)/2 & 3.4(1+4/9)/2 - E \end{array}\right|$
(b)$\left|\begin{array}{cc} 3.4(1+4/9)/2 - E & 3.4(1-4/9)/2 \\ 3.4(1-4/9)/2 & 3.4(1-4/9)/2 - E \end{array}\right|$
(c)$\left|\begin{array}{cc} 3.4(1+4/9)/2 - E & 3.4(1+4/9)/2 \\ 3.4(1-4/9)/2 & 3.4(1-4/9)/2 - E \end{array}\right|$
(d)$\left|\begin{array}{cc} 3.4(1+4/9)/2 - E & 0 \\ 0 & 3.4(1+4/9)/2 - E \end{array}\right|$
Answer
Answer (as printed): A
Explanation
No explanation was printed for this question.

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