Question

For ${\mathrm{He}}^{+}$the electron is in orbit with energy equal to $3.4\mathrm{eV}$. The azimuthal quantum number for that orbit is $2$ and the magnetic quantum number is $0$. Then which of the following is/are correct?

• A. The subshell is $4\mathrm{d}$.
• B. The number of angular nodes in it is $2$.
• C. The numbers of radial nodes in it is $3$
• D. The nuclear charge experienced in $\mathrm{n}=4$ is $2\mathrm{e}$ less than that in $\mathrm{n}=1$, where $\mathrm{e}$ is the electric charge.

Answer 1

Answer: (A), (B)

The number of angular nodes in it is $2$.

Explanation for the correct option:

Step-1

The energy of electron in Bohr's orbit:

$\mathrm{E}=\frac{{\mathrm{E}}_{\mathrm{o}}\times {\mathrm{Z}}^2}{{\mathrm{n}}^2}$

where, $\mathrm{Z}$ is the atomic no. of Hydrogen like species

$\mathrm{n}$ is the principal quantum number

$\mathrm{l}$ is azimuthal quantum number

Step 2: Putting all the variables in above equation:

$3.4=13.6\times \left(4/{\mathrm{n}}^2\right)$

After solving, we get:

$$\begin{gathered} \mathrm{n}=4 \\ \mathrm{l}=2 \end{gathered}$$

Step-3:

Therefore, from the above calculation it is clear that:

(A) The subshell is $4\mathrm{d}$

(B) Number of angular nodes $=\mathrm{l}$

Number of angular nodes is $2$

Explanation for the incorrect option:

(C) Number of radial nodes $\mathrm{n}-\mathrm{l}-1$

Number of radial nodes $$\begin{gathered} =4-2-1 \\ =1 \end{gathered}$$

Therefore, option (C) is incorrect.

(D) Nuclear charge would be the same hence, option (D) is incorrect

Therefore, the correct answer options are (A), (B).