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What is the wavelength of light in nm that is emitted when an excited electron in the hydrogen atom falls from n=5 to n=2?

a) 410 nm
b) 656 nm
c) 486 nm
d) 6561 nm

User Parobay
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1 Answer

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Final answer:

The wavelength of light emitted when an excited electron in the hydrogen atom falls from n=5 to n=2 is approximately 434 nm.

Step-by-step explanation:

The wavelength of light emitted when an excited electron in the hydrogen atom falls from n=5 to n=2 can be calculated using the Rydberg formula:

1/λ = R (1/nf^2 - 1/ni^2)

where λ is the wavelength, R is the Rydberg constant (approximately 1.097 x 10^7 m^-1), and nf and ni are the final and initial energy levels respectively.

Substituting the values nf=2 and ni=5 into the formula, we get:

1/λ = 1.097 x 10^7 (1/2^2 - 1/5^2)

Simplifying the equation gives us:

1/λ = 1.097 x 10^7 (1/4 - 1/25)

1/λ = 1.097 x 10^7 (21/100)

1/λ = 1.097 x 10^7 x 0.21

1/λ = 2.3027 x 10^6

Using the reciprocal of both sides:

λ = 4.34 x 10^-7 m

Converting to nanometers (nm), the wavelength is approximately 434 nm.

User Dosdos
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