The wavefunction of a 1 – D harmonic oscillator between x=+∞ and x=-∞ is given by ψ(x)=N(2x²-1)e-x²/2. The value of N that normalizes the function ψ(x) is: (Given ∫-∞+∞x²ⁿe-x² dx=(1·3·5…(2n-1))/(2n)√(π))

The wavefunction of a 1 – D harmonic oscillator between x=+∞ and x=-∞ is given by ψ(x)=N(2x²-1)e-x²/2. The… · ABC26PH0299

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

The wavefunction of a 1 – D harmonic oscillator between $x=+\infty$ and $x=-\infty$ is given by $\psi(x)=N(2x^2-1)e^{-x^2/2}$. The value of $N$ that normalizes the function $\psi(x)$ is: $\left(\text{Given }\int_{-\infty}^{+\infty}x^{2n}e^{-x^2}\,dx=\frac{1\cdot3\cdot5\cdots(2n-1)}{2^n}\sqrt{\pi}\right)$

(a)$\left(\frac{1}{8\sqrt{\pi}}\right)^{\frac{1}{2}}$

(b)$\left(\frac{1}{3\sqrt{\pi}}\right)^{\frac{1}{2}}$

(c)$\left(\frac{1}{2\sqrt{\pi}}\right)^{\frac{1}{2}}$

(d)$\left(\frac{1}{4\sqrt{\pi}}\right)^{\frac{1}{2}}$

Answer

Answer (as printed): C

Explanation

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