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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

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 Correct Answer $$\frac{2}{{{a^2}}}$$

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A particle of mass m is confined in the potential
Quantum Mechanics mcq question image
\[V\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {\frac{1}{2}m{\omega ^2}{x^2},}&{{\text{for }}x Let the wave function of the particle be given by $$\psi \left( x \right) = - \frac{1}{{\sqrt 5 }}{\psi _0} + \frac{2}{{\sqrt 5 }}{\psi _1}$$
where $${\psi _0}$$ and $${\psi _1}$$ are the eigen functions of the ground state and the first excited slate respectively. The expectation value of the energy is
A particle is in the normalized state $$\left| \psi \right\rangle $$ which is a superposition of the energy eigen states $$\left| {{E_0} = 10\,eV} \right\rangle $$   and $$\left| {{E_1} = 30\,eV} \right\rangle .$$   The average value of energy of the particle in the state $$\left| \psi \right\rangle $$ is 20 eV. The state $$\left| \psi \right\rangle $$ is given by
The expectation value of the z coordinate, (z), in the ground state of the hydrogen atom (wave function: $${\psi _{100}}\left( r \right) = A{e^{ - \frac{r}{{{a_0}}}}},$$    where A is the normalization constant and $${a_0}$$ is the Bohr radius), is
If a + b + c + d = 4, then find the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
If a + b + c + d = 4, then the value of $$\frac{1}{{\left( {1 - a} \right)\left( {1 - b} \right)\left( {1 - c} \right)}}$$     + $$\frac{1}{{\left( {1 - b} \right)\left( {1 - c} \right)\left( {1 - d} \right)}}$$     + $$\frac{1}{{\left( {1 - c} \right)\left( {1 - d} \right)\left( {1 - a} \right)}}$$     + $$\frac{1}{{\left( {1 - d} \right)\left( {1 - a} \right)\left( {1 - b} \right)}}$$     is?
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The value of the expression $$\frac{{{{\left( {a - b} \right)}^2}}}{{\left( {b - c} \right)\left( {c - a} \right)}} + $$   $$\frac{{{{\left( {b - c} \right)}^2}}}{{\left( {a - b} \right)\left( {c - a} \right)}} + $$    $$\frac{{{{\left( {c - a} \right)}^2}}}{{\left( {a - b} \right)\left( {b - c} \right)}}$$   = ?
The Hamiltonian of a particle is given by $$H = \frac{{{p^2}}}{{2m}} + V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right) + \phi \left( { + \left| {\overrightarrow {\bf{r}} } \right|} \right)\overrightarrow {\bf{L}} .\overrightarrow {\bf{S}} ,$$        where $$\overrightarrow {\bf{S}} $$ is the spin, $$V\left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  and $$\phi \left( {\left| {\overrightarrow {\bf{r}} } \right|} \right)$$  are potential functions and $$\overrightarrow {\bf{L}} \left( { = \overrightarrow {\bf{r}} \times \overrightarrow {\bf{p}} } \right)$$   is the angular momentum. The Hamiltonian does not commute with
A system in a normalized state $$\left| \psi \right\rangle = {c_1}\left| {{\alpha _1}} \right\rangle + {c_2}\left| {{\alpha _2}} \right\rangle $$    with $$\left| {{\alpha _1}} \right\rangle $$ and $$\left| {{\alpha _2}} \right\rangle $$ representing two different eigen states of the system requires that the constants c1 and c2 must satisfy the condition
$$\frac{{{{\left( {4.53 - 3.07} \right)}^2}}}{{\left( {3.07 - 2.15} \right)\left( {2.15 - 4.53} \right)}} + \, $$     $$\frac{{{{\left( {3.07 - 2.15} \right)}^2}}}{{\left( {2.15 - 4.53} \right)\left( {4.53 - 3.07} \right)}} + \,\, $$     $$\frac{{{{\left( {2.15 - 4.53} \right)}^2}}}{{\left( {4.53 - 3.07} \right)\left( {3.07 - 2.15} \right)}}$$     is simplified to :