Question

The wave function, ${\psi }_{n,l,m_{\ell }}$ is a mathematical function whose value depends upon spherical polar coordinates (r, $\theta ,\varphi$ ) of the electron and characterized by the quantum numbers $n,\ell$ and $m_l.$ Here $r$ is distance from nucleus, $\theta$ is colatitude and $\varphi$ is azimuth. In the mathematical functions given in the Table, $Z$ is atomic number and $a_0$ is Bohr radius.

Column 1	Column 2	Column 3
(I) 1s orbital	(i) ${\psi }_{n,l,m_l}\propto {\left(\frac{Z}{a_e}\right)}^{\frac{3}{2}}{e}^{-\left(\frac{z}{a_0}\right)}$	(p)
(II) 2s orbital	(ii) one radial node	(Q) Probability density at nucleus $\propto \frac{1}{a_0^3}$
(III) $2p_z$ orbital	(iii) ${\psi }_{n,l,m_l}\propto {\left(\frac{Z}{a_o}\right)}^{\frac{5}{2}}r{e}^{-\left(\frac{z}{2a_0}\right)}\cos \theta$	(R) Probability density is maximum at nucleus
(IV) $3d_z^2$ Orbital	(iv) xy-plane is a nodal plane	(S) Energy needed to excite electron from $n=2$ state to $n=4$ state is $\frac{27}{32}$ times the energy needed to excite electron from $n=2$ state to $n=6$ state

For hydrogen atom, the only CORRECT combination is

• A. (I) (ii) (S)
• B. (IV) (iv) (R)
• C. (II) (ii) (P)
• D. (III) (iii) (P)

Answer 1

Answer: (D)

$\frac{E_4\to 2}{E_6\to 2}=\frac{\left[\frac{1}{2^2}-\frac{1}{4^2}\right]}{\left[\frac{1}{2^2}-\frac{1}{6^2}\right]}=\frac{27}{32}$