Question

The radial wave function of an orbital is $2{\left( {\dfrac{Z}{{{a_0}}}} \right)^{\dfrac{3}{2}}}{e^{\dfrac{{Zr}}{{ - {a_0}}}}}$ for H-atom. The wave function is for-(notations have their usual meaning)
A.$1s$
B. $2s$
C. $2p$
D. $3p$

Answer 1

Hint: The wave function for Hydrogen atoms depends upon three variables one of which is $\theta $ which gives the electron's angular momentum.

Step-by-Step Solution: We are given the radial wave function of an orbital = $2{\left( {\dfrac{Z}{{{a_0}}}} \right)^{\dfrac{3}{2}}}{e^{\dfrac{{Zr}}{{ - {a_0}}}}}$
Where r=radius in atomic units, Z= the effective nuclear charge for that orbital in that atom
e=$2.718$. We have to find the wave function.
The wave function gives crucial information about the electron’s energy, angular momentum and orbital orientation. The wave function for a hydrogen atom depends upon three variables one of which is $\theta $ which gives the electron's angular momentum.
And the radial wave function of an orbital = $2{\left( {\dfrac{Z}{{{a_0}}}} \right)^{\dfrac{3}{2}}}{e^{\dfrac{{Zr}}{{ - {a_0}}}}}$ is for $1s$ orbital because it gives
$p(\theta ) = \dfrac{1}{{\sqrt 2 }}$ .

Hence the correct answer is A.

Additional Information: Wave function is a mathematical function represented by φ(psi). Wave function can also be expressed as the product of radial wave function R and angular wave function. The R radial wave function of an atom depends upon the atomic radius only while angular function depends only on direction thus describes the shape of an orbital. Thus we can write-
$ \Rightarrow $ φ=R×Y

Note: The wave function can be calculated exactly only for atoms with one electron like hydrogen,$H{e^ + }$ and other atoms having only one electron. This means that wave function is possible only for said one-electron system which is also described as hydrogenic. Hydrogenic means ‘like hydrogen’. Wave function can be obtained by solving the Schrödinger equation. It explains why there is a single $1s$orbital, why there are three $2p$ orbitals and so on.