(a) Neglecting reduced-mass effects, what optical transition in the (\text{He}^{+}) spectrum would have the same wavelength as the first Lyman transition of hydrogen ((n=2) to (n=1))? (b) What is the second ionization energy of (\text{He})? © What is the radius of the first Bohr orbit for (\text{He}^{+})? Assume that the ionization energy ((\hat{v})) of deuterium is (R).

Let's respond to each component of the query separately:
(a) What optical transition in the (He+) spectrum, ignoring reduced-mass effects, would have the same wavelength as the first Lyman transition of hydrogen (n=2 to n=1)?
The formula for the wavelength of a spectral line can be used to determine the wavelength of a transition in atoms that resemble hydrogen (like He+):
1 / λ = R_H * (1/n₁² - 1/n₂²)
Where:
    The transition's wavelength is •.
    The Rydberg constant for hydrogen is R_H.
    The major quantum numbers for the two energy levels are n1 for the beginning energy level and n2 for the final energy level.
We have n1 = 2 and n2 = 1 for the first Lyman transition in hydrogen (n=2 to n=1).
He+ (the helium ion) only has one electron at this time. As a result, it has a comparable electron configuration to hydrogen, and the equivalent transition can be determined using the same formula and the hydrogen Rydberg constant (R_H).

1 / _He+ = R_H * (1/n12-1/n22) = R_H * (1/22-1/12) = R_H * (1/4-1) = R_H * (3/4)

The wavelength of the He+ transition, which corresponds to the first Lyman transition in hydrogen, is therefore 4/3 times that of the hydrogen transition.

(b) What is the second energy at which (He) ionises?

The energy needed to expel the second electron from a helium atom (He2+ He3+ + e) is known as the second ionisation energy of helium (He). We may compute it using the idea of ionisation energy and the knowledge that deuterium (D2) has an ionisation energy of R.

The equation En = n2*Z2/RH can be used to calculate the ionisation energy for any atom that resembles a hydrogen atom.​​​​

Where:

    The energy of the nth energy level is called En.
    The atomic number Z is the quantity of protons in a certain nucleus.
    The hydrogen-specific Rydberg constant is RH.
    The fundamental quantum number is n.
Helium (He) has two protons, hence Z=2 is the ionisation energy for helium at its second ionisation. When n=1n=1 for He2+, we want to determine the energy.

En=1=−22RH12=−4RH

Now that deuterium's (D2) ionisation energy is RR, we may construct the following equation:

En=1(He²+)=En=1(D2)

−4RH=−R

We can now determine R, the deuterium ionisation energy:

R=4RH

As a result, deuterium's (D2) second ionisation energy is 4 times greater than that of helium (He).

(c) What is the first Bohr orbit for (He+)'s radius?

Bohr's formula can be used to determine the radius of the initial Bohr orbit for any element that is similar to hydrogen:

r=n2⋅h2/4π2⋅m⋅e2⋅Z

Where:

    The orbit's radius is r.
    The fundamental quantum number, or n, is 1 in this case for the initial Bohr orbit.
    The Planck constant is h.
    When reduced-mass effects are taken into account, the electron's mass, m, is the reduced mass of the electron and nucleus system, or m.
    The fundamental charge is e.
    The atomic number Z is the quantity of protons in a certain nucleus.

Since the helium ion He+ has two protons, Z=2.

Put in the following known values:

r=1^2⋅h^2/4π^2⋅m⋅e^2⋅

Using the mass of the electron and disregarding reduced-mass effects, we have:

m=me

You may now calculate r by substituting the known values for h, me, and e.

The Bohr's formula for the orbital radius can be used to get the exact value of the radius of the first Bohr orbit for the helium ion (Z=2):

r=n^2⋅h^2/4π^2⋅m⋅e^2⋅Z

Where:

    The primary quantum number is n; during the first orbit, n equals 1.
    The Planck constant, h, is equal to 6.626070151034 m2 kg/s.
    The electron weighs 9.10938356 x 10^-31 kilograms, or m.
    The fundamental charge is e (1.60217663 1019 C).
    Z is the atomic number, and for helium, Z=2.

Now enter the following values into the formula:
Let's now determine the sum in numbers:

r≈5.2917721092×10^−11m

Therefore, 5.2917721092 x 10^-11 metres is roughly the numerical value for the radius of the first Bohr orbit for the helium ion (Z=2).