1. Use the hydrogen-like energy formula

For a hydrogen-like species, $E_n=-13.6\,\frac{Z^2}{n^2}\text{ eV}$ where $Z$ is the atomic number.

For $He^+$, we have $Z=2$. So, $E_n=-13.6\,\frac{(2)^2}{n^2}=-54.4\,\frac{1}{n^2}\text{ eV}$

Given energy of the state: $E_n=-3.4\text{ eV}$ Thus, $-\frac{54.4}{n^2}=-3.4$ $n^2=\frac{54.4}{3.4}=16$ $n=4$

So the electron is in the $n=4$ shell.

2. Interpret the azimuthal quantum number

Given: $l=2$

The subshell corresponding to $l=2$ is $d$. Hence the state is: $4d$

So Option A is true.

3. Check the magnetic quantum number

Given: $m_l=0$ This is allowed for $l=2$, since $m_l=-2,-1,0,+1,+2$ So the state is consistent with $4d$.

4. Check the effective nuclear charge statement

Since $He^+$ is a one-electron species, there is no electron-electron repulsion or shielding by other electrons. Therefore the electron experiences the full nuclear charge: $+2e$ not less than $2e$.

So Option B is false.

5. Find the number of angular nodes

Number of angular nodes is equal to $l$. Thus, $\text{angular nodes}=l=2$

So Option C is true.

6. Find the number of radial nodes

Number of radial nodes is given by: $n-l-1$ Substituting $n=4$, $l=2$: $4-2-1=1$

So the state has 1 radial node, not 3.

Therefore Option D is false.

7. Final evaluation of options

• A: True
• B: False
• C: True
• D: False

Hence the correct statements are: $\boxed{A,\ C}$