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.

8.56*10^6

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

1/λ = R(1/n₁² - 1/n₂²)

Where:
- λ is the wavelength of light emitted
- R is the Rydberg constant (approximately 1.097 x 10^7 m^-1)
- n₁ and n₂ are the principal quantum numbers of the initial and final energy levels, respectively.

In this case, we need to consider transitions between n=2, n=3, and n=4. Let's calculate the lowest wavenumber for each transition.

1. Transition from n=2 to n=3:
Using the Rydberg formula, we have:
1/λ₁ = R(1/2² - 1/3²)

Simplifying:
1/λ₁ = R(1/4 - 1/9)
1/λ₁ = R(5/36)

Now, solve for λ₁:
λ₁ = 36/(5R)

Convert the wavelength to wavenumber:
ν₁ = 1/λ₁

2. Transition from n=3 to n=4:
Using the Rydberg formula, we have:
1/λ₂ = R(1/3² - 1/4²)

Simplifying:
1/λ₂ = R(1/9 - 1/16)
1/λ₂ = R(7/144)

Now, solve for λ₂:
λ₂ = 144/(7R)

Convert the wavelength to wavenumber:
ν₂ = 1/λ₂

To get the lowest wavenumber, we need to compare ν₁ and ν₂ and select the smaller value.

3. Transition from n=4 to infinity:
The transition from n=4 to infinity represents the highest energy transition possible. It corresponds to the Lyman series limit. In this case, the wavenumber would be the same as the ionization energy, which is beyond the scope of this calculation.

Therefore, the lowest wavenumber νˉ (in inverse meters) would be the smaller value between ν₁ and ν₂. You can calculate these values using the above formulas and the given Rydberg constant.