You're asking about the emission of light when an electron transitions from the n=7 to n=3 energy level in an atom. To figure out the wavelength, we need to use the Rydberg formula.

The Rydberg Formula

The Rydberg formula is:

1/λ = R * (1/n₁² - 1/n₂²)

where:

* λ is the wavelength of the emitted light (in meters)

* R is the Rydberg constant (approximately 1.097 × 10⁷ m⁻¹)

* n₁ is the lower energy level (in this case, n₁ = 3)

* n₂ is the higher energy level (in this case, n₂ = 7)

Calculations

1. Plug in the values:

1/λ = 1.097 × 10⁷ m⁻¹ * (1/3² - 1/7²)

2. Solve for 1/λ:

1/λ ≈ 9.14 × 10⁵ m⁻¹

3. Solve for λ (wavelength):

λ ≈ 1.09 × 10⁻⁶ meters

4. Convert to nanometers (nm):

λ ≈ 1090 nm

Conclusion

The astrophysicist should look for light with a wavelength of approximately 1090 nanometers to detect this transition. This wavelength falls within the near-infrared portion of the electromagnetic spectrum.