An atomic state of hydrogen is represented by following wave function $$\psi \left( {r,\,\theta ,\,\phi } \right) = \frac{1}{{\sqrt 2 }}{\left( {\frac{1}{{{a_0}}}} \right)^{\frac{3}{2}}}\left( {1 - \frac{r}{{2{a_0}}}} \right){e^{\frac{{ - r}}{{2{a_0}}}}}\cos \theta $$ where, a0 is a constant. The quantum numbers of the state are
A. l = 0, m = 0, n = 1
B. l = 1, m = 1, n = 2
C. l = 1, m = 0, n = 2
D. l = 2, m = 0, n = 3
Answer: Option B
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. e-ax2 (e-ax1 - e-ax2)
A. 0.75
B. 0.50
C. 0.35
D. 0.25
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