A monochromatic light beam with energy 204eV is used to excite the Be+3 (H –like atom with Z=4) from its ground state to some excited state m. Neglect the finite nuclear mass correction.

a) Find the quantum number m of this excited state.
b) Following this transition the excited Be+3 atom emits a spectral line at 30.42nm as a result of the transition from this excited state m to some other lower state n. Find the quantum number of this new state.
c) Calculate the orbital radius of the electron in the state n.

(a) For BeIII with Z=4, use the Rydberg equation for the energy levels. They will be 4^2 = 16 times more widely separated than for hydrogen.

E(m) = 13.58 eV*16/n^2 = 217.3 /m^2 eV

For a transition from the ground state (m=1),

204 eV = 217.3/1 - 217.3/m^2
217.3/m^2 = 16.4

m = 4 (probably) There may be some accuracy issues with the value of the Rydberg that I used.

(b) Convert the wavelength to photon energy in eV. Then use the Rydberg equation to compute n.

E = h*c/(wavelength) = ____ eV

E = 204 eV - R/n^2 = 204 - 217.3/n^2
Solve for n

(c) Bohr orbit radius = ao * n^2/Z
where ao is the hydrogen ground state Bohr radius

To find the quantum number m of the excited state, we need to use the formula for energy of a hydrogen-like atom:

En = - (m_e^4 * Z^2) / (8 * h^2 * ε_0^2 * n^2)

where:
En is the energy of the excited state
m_e is the electron mass
Z is the atomic number
h is the Planck's constant
ε_0 is the vacuum permittivity
n is the principal quantum number

In this case, Z = 4 (Be+3), and the energy of the monochromatic light beam is given as 204 eV. We need to convert it to Joules by multiplying with the elementary charge (e) and converting it to Joules:

E = 204 * 1.6e-19 Joules

Now, we can solve for n using the equation:

E = - (m_e^4 * Z^2) / (8 * h^2 * ε_0^2 * n^2)

Rearranging the equation:

n^2 = - (m_e^4 * Z^2) / (8 * h^2 * ε_0^2 * E)

Solving for n:

n = sqrt(- (m_e^4 * Z^2) / (8 * h^2 * ε_0^2 * E))

Now, calculate the value of n for the given equation. Plugging in the values:

n = sqrt(- (9.11e-31 kg)^4 * (4^2) / (8 * (6.63e-34 J.s)^2 * (8.85e-12 F/m)^2 * 204 * 1.6e-19 J))

Solving this equation will give you the value of n, which corresponds to the quantum number m for the excited state of Be+3.

To find the quantum number of the new state (lower state), we can use the formula for wavelength of a spectral line emitted during a transition:

1/λ = R * (Z^2 / n^2 - Z^2 / m^2)

where:
λ is the wavelength of the emitted line
R is the Rydberg constant (1.097e7 m^-1)
Z is the atomic number
n is the initial principal quantum number
m is the final principal quantum number

In this case, Z = 4 (Be+3), and the wavelength is given as 30.42 nm. We need to convert it to meters:

λ = 30.42e-9 meters

Now, we can solve for m using the equation:

1/λ = R * (Z^2 / n^2 - Z^2 / m^2)

Rearranging the equation:

1/m^2 = Z^2 / n^2 - (λ / R) * Z^2

Solving for m:

m^2 = n^2 / (1 - (λ / R) * Z^2 / n^2)

Now, calculate the value of m for the given equation. Plugging in the values:

m^2 = n^2 / (1 - (30.42e-9 m) / (1.097e7 m^-1) * (4^2) / n^2)

Solving this equation will give you the value of m, which corresponds to the quantum number of the new state.

To calculate the orbital radius of the electron in the state n, we can use the formula for the orbital radius:

r = a₀ * n^2 / Z

where:
r is the orbital radius
a₀ is the Bohr radius (0.529e-10 meters)
n is the principal quantum number
Z is the atomic number

In this case, Z = 4 (Be+3), and we have already found the value of n in the previous parts. Plugging in the values:

r = (0.529e-10 m) * n^2 / 4

Solving this equation will give you the value of r, which corresponds to the orbital radius of the electron in the state n.