The normalized ground state, wave function of a hydrogen atom is given $$\psi \left( r \right) = \frac{1}{{\sqrt {4\pi } }}\frac{2}{{{a^{\frac{3}{2}}}}} - {e^{ - \frac{r}{a}}},$$ where a is the Bohr radius and r is the distance of the electron from the nucleus located at the origin. The expectation value $$\left\langle {\frac{1}{{{r^2}}}} \right\rangle $$ is
A. $$\frac{{8\pi }}{{{a^2}}}$$
B. $$\frac{{4\pi }}{{{a^2}}}$$
C. $$\frac{4}{{{a^2}}}$$
D. $$\frac{2}{{{a^2}}}$$
Answer: Option D
Related Questions on Quantum Mechanics
A. In the ground state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half
B. In the first excited state, the probability of finding the particle in the interval $$\left( {\frac{L}{4},\,\frac{{3L}}{4}} \right)$$ is half This also holds for states with n = 4, 6, 8, . . . .
C. For an arbitrary state $$\left| \psi \right\rangle ,$$ the probability of finding the particle in the left half of the well is half
D. In the ground state, the particle has a definite momentum
A. (e-ax1 - e-ax2)
B. a(e-ax1 - e-ax2)
C. e-ax2 (e-ax1 - e-ax2)
D. None of the above

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