Answer
See the detailed answer below.
Work Step by Step
We know that an ion with a single electron orbiting $Z$ protons in the nucleus is called a hydrogen-like ion, where $Z$ is the number of protons in the nucleus. Hence, $\rm He^+$, with one electron circling a $Z = 2$ nucleus is a hydrogen-like ion.
And since any hydrogen-like ion is simply a variation on the Bohr hydrogen atom, we can use the following formulas.
To find the radius of the electron’s orbit,
$$r_n=\dfrac{n^2a_B}{Z}\tag 1$$
To find the electron’s speed,
$$v_n=Z\dfrac{v_1}{n}\tag 2$$
And to find the energy of the atom,
$$E_n= \dfrac{- 13.6Z^2}{n^2} \tag 3$$
For the first three stationary states; where $n=1,2,3$
The radius of the electron’s orbit at $n=1$, using (1);
$$r_1=\dfrac{(1^2) (0.053)}{2} =\color{red}{\bf 0.0265 }\;\rm nm$$
The radius of the electron’s orbit at $n=2$;
$$r_2=\dfrac{(2^2) (0.053)}{2} =\color{red}{\bf 0.106}\;\rm nm$$
The radius of the electron’s orbit at $n=3$;
$$r_3=\dfrac{(3^2) (0.053)}{2} =\color{red}{\bf 0.2385}\;\rm nm$$
The electron’s speed at $n=1$, using (2);
$$v_1=(2)\dfrac{(2.19\times 10^6)}{(1)} =\color{red}{\bf 4.38\times 10^6}\;\rm m/s$$
The electron’s speed at $n=2$;
$$v_2=(2)\dfrac{(2.19\times 10^6)}{(2)} =\color{red}{\bf 2.19\times 10^6}\;\rm m/s$$
The electron’s speed at $n=3$;
$$v_3=(2)\dfrac{(2.19\times 10^6)}{(3)} =\color{red}{\bf 1.46\times 10^6}\;\rm m/s$$
The energy of the atom at $n=1$, using (3);
$$E_1= \dfrac{- 13.6(2)^2}{(1)^2} =\color{red}{\bf -54.4}\;\rm eV $$
The energy of the atom at $n=2$,
$$E_2= \dfrac{- 13.6(2)^2}{(2)^2} =\color{red}{\bf -13.6}\;\rm eV $$
The energy of the atom at $n=3$,
$$E_3= \dfrac{- 13.6(2)^2}{(3)^2} =\color{red}{\bf -6.04}\;\rm eV $$
