Schrodinger Equation

Q.

The wave function $\psi$ in the Schrodinger wave equation represents

1. Probability of the electron

2. Amplitude of the wave

3. Frequency of the wave

4. Speed of the wave

Q. The complete wave function of an orbital of hydrogen is given as $\psi =\frac{1}{4a_0\sqrt{2\pi a_0}}(2-\frac{r}{a_0})e^{\frac{-r}{2a_0}}$,
($a_0$ is first Bohr radius)
Which of the following option(s) is/are correct?

1. Orbital may be 3p
2. Angular node is at $2a_0$ from nucleus
3. Ratio of density of probability of finding electron at distance $a_0$ to nucleus is $\frac{e^{-1}}{4}$
4. Orbital has a total of one node

Q. As compared to the $1s$ electron of $H-$ atom in ground state, which of the following properties appear(s) in the radial probability density of electron of $H-$ atom in first excited state ?

1. Spherical node appears
2. Electron probability density is highest in the vicinity of the nucleus
3. Probability density drops to zero after maximum probability is reached
4. Probability density rises to second highest value

Q. Consider following Schrodinger wave equation for an orbital of hydrogen atom.
$(a_0$ is first Bohr radius)
$\Psi =\frac{\sqrt{2r}}{81a_0^2\sqrt{\pi a_0}}(6-\frac{r}{a_0})e^{\frac{-r}{3a_0}}cos\theta$

$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(I\right)& \text{Which orbital is this?}& \left(P\right)& 1\\ \left(II\right)& \text{Number of total node(s)}& \left(Q\right)& 2\\ \left(III\right)& \text{Nodal plane}& \left(R\right)& 3p_z\\ \left(IV\right)& \text{Number of radial node(s)}& \left(S\right)& 3p_x\\ & & \left(T\right)& \text{XY plane}\\ & & \left(U\right)& \text{YZ plane}\end{array}$

Match the correct combination considering List-I and List-II

1. (I), (R) and (II), (P)
   
2. (I), (S) and (II), (P)
3. (I), (R) and (II), (Q)
4. (I), (S) and (II), (Q)