What is the wavelength (in nanometers) of light emitted from a hydrogen atom when an electron falls from the n = 2 to n = 1 orbital?

I got this answer and i wanted to make sure it is right

121.7

Thank you

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To calculate the wavelength of light emitted when an electron falls from one energy level to another in a hydrogen atom, you can use the Rydberg formula:

1/λ = R_H * (1/n_f^2 - 1/n_i^2)

Where:
- λ is the wavelength of light emitted
- R_H is the Rydberg constant for hydrogen (approximately 1.097 x 10^7 m^-1)
- n_f is the final energy level (n = 1, in this case)
- n_i is the initial energy level (n = 2, in this case)

Let's plug in the values and solve for λ:

1/λ = (1.097 x 10^7 m^-1) * (1/1^2 - 1/2^2)

1/λ = (1.097 x 10^7 m^-1) * (1 - 1/4)

1/λ = (1.097 x 10^7 m^-1) * (3/4)

1/λ = 8.2275 x 10^6 m^-1

Now, we can find the wavelength by taking the reciprocal of both sides:

λ = 1 / (8.2275 x 10^6 m^-1)

λ ≈ 1.2143 x 10^-7 m

To convert this value to nanometers (nm), we multiply by 10^9:

λ ≈ 121.43 nm

Therefore, the correct wavelength of light emitted when an electron falls from the n = 2 to n = 1 orbital in a hydrogen atom is approximately 121.43 nm.