for the emission spectrum of Be3+, calculate the lowest wavenumber νˉ (in inverse meters) of light produced by electron transitions between n=2, n=3, and n=4.

Question

for the emission spectrum of Be3+, calculate the lowest wavenumber νˉ (in inverse meters) of light produced by electron transitions between n=2, n=3, and n=4.

Answer 1

An electronic transition occurs between two energy levels. Why have you listed three?

For Be3+, you can use the Rydberg formaula for energy levels, with Z = 4. The nuclear charge is +4e, and there is one orbiting electron

Perhaps what they want is the wave number for the transitions 3->2, 4->3 and 5->4.

These will be 16 times the corresponding transitions for hydrogen

Answer 2

Hello drwls,

could you be more specific about the forumla and how we can solve this problem ?

Answer 3

Hi DRWLS, please be bound by MIT EDX honor code or else you will cause trouble for everyone.

Thank you for your kind corporation.

Answer 4

Jiskha is not part of the MITx platform!!! How the honor code of MITx could be applied to an user of other system?

Answer 5

good point

Answer 6

To calculate the lowest wavenumber (νˉ) of light produced by electron transitions for the emission spectrum of Be3+, we need to use the Rydberg formula. The Rydberg formula for the wavenumber (νˉ) is given by:

1/λ = R * (Z²/n²)

Where:
λ is the wavelength of the emitted light,
R is the Rydberg constant,
Z is the atomic number,
and n is the principal quantum number.

For Be3+, Z = 4, since it has 4 protons in its nucleus (atomic number = number of protons).

To find the lowest wavenumber, we need to calculate the wavelength (λ) using the Rydberg formula and then convert it to wavenumber (νˉ).

For the first transition between n = 2 to n = 1, we use n = 2 and n = 1 in the formula:

1/λ = R * (Z²/n²)
1/λ₁₂ = R * (4²/2²) = 4R

For the second transition between n = 3 to n = 1, we use n = 3 and n = 1 in the formula:

1/λ₃₁ = R * (4²/3²) = 16R/9

For the third transition between n = 4 to n = 1, we use n = 4 and n = 1 in the formula:

1/λ₄₁ = R * (4²/4²) = R

Now, we convert the calculated wavelengths to wavenumbers (νˉ) by taking the reciprocal of the wavelength (λ):

νˉ₁₂ = 1/λ₁₂ = 1/(4R)
νˉ₃₁ = 1/λ₃₁ = 9/(16R)
νˉ₄₁ = 1/λ₄₁ = 1/R

Therefore, the lowest wavenumber (νˉ) of light produced by electron transitions between n = 2, n = 3, and n = 4 for Be3+ are:

νˉ₁₂ = 1/(4R)
νˉ₃₁ = 9/(16R)
νˉ₄₁ = 1/R

Please note that the actual values of the Rydberg constant (R) and conversions to wavenumbers depend on the units used.