Question

The ground state energy of a hydrogen atom is \[ - 13.6eV\]. Consider an electronic state of \[\psi \] of \[H{e^ + }\] whose energy, azimuthal quantum number and magnetic number are \[ - 3.4eV\] , \[2\] and \[0\] respectively. Which of the statements is or are true for the state \[\psi \]?
A. It is \[4d\] state
B. The nuclear charge experienced by the electron in this state is less than \[2e\] , where \[e\] is the magnitude of electric charge.
C. It has \[3\] radial nodes.
D. It has \[2\] angular nodes

Answer 1

Hint: Energy is given so that we can find the particular orbit number \[n\] and azimuthal quantum number \[l\] is also given hence we can find orbital state . Angular node is the same as that of azimuthal quantum number. From \[n\] and \[l\] radial nodes can also be calculated.
Formula used: Energy \[ = \dfrac{{ - 13.6 \times {Z^2}}}{{{n^2}}}\]
where \[Z\] is the atomic number of particular element
where \[n\] is the principal quantum number (particular orbit in which the atom belongs to).

Complete answer:
Energy given here is \[ - 3.4eV\]
To find energy in a particular orbit of hydrogen atom is given by the formula \[ = \dfrac{{ - 13.6 \times {Z^2}}}{{{n^2}}}\]
Therefore \[ - 3.4eV = \dfrac{{ - 13.6 \times {Z^2}}}{{{n^2}}}\]
where \[Z\] is the atomic number of helium \[ = 2\]
From here we need to find \[n\] from \[{n^2}\]
On substituting the values \[Z\] in the above formula we get ,
\[{n^2} = \dfrac{{13.6 \times {2^2}}}{{3.4}}\]
\[{n^2} = 16\]
Therefore \[n = 4\]
Here azimuthal quantum number \[l\] is given by \[2\] which means the atom belongs to sub shell \[d\]
That is it belongs to \[4d\] state and hence option A is correct.
Next we need to look onto angular nodes. The point to remember here is that the number of angular nodes is equal to azimuthal quantum number \[l\] which is given in question itself which is \[2\] .
Therefore number of angular node is \[2\]
Hence option D is also correct.
Next we have to find the number of radial nodes. Here we need to recall an easy equation which is as follows,
Number of radial nodes \[ = n - l - 1\]
We already calculated \[n\] and \[l\] above which is \[4\] and \[2\] respectively.
On substituting the values on the above equation we get ,
\[4 - 2 - 1 = 1\]
 Hence option C is incorrect which is stating about having three radial nodes.
Finally we have to know that \[H{e^ + }\] is having a single electronic system. Therefore there is no chance of experiencing a shielding effect or in other words no nuclear charge. If there is a shielding effect then there will be a net nuclear charge .
Hence it proves option B is incorrect which is describing net nuclear charge experienced by the electron.

Note:The probability of finding an electron is said to be zero in case of radial nodes. Angular nodes are equal to the value of azimuthal quantum number, and since the value for azimuthal quantum number for d orbital is two, the value for angular nodes also has to be two. The number of nodes increases when the principal quantum number increases.