Question

Calculate the wavelength (in nm) of a photon emitted when an electron

transitions from n = 3 to n = 2 in a hydrogen atom.
O 730 nm
O 657 nm
O 846 nm
O 308 nm
O 591 nm

Answer 1

Using the Rydberg formula, the wavelength of the photon emitted when an electron transitions from n = 3 to n = 2 in a hydrogen atom is approximately 656.5 nm, which is rounded to 657 nm.

Step-by-step explanation:

To calculate the wavelength of a photon emitted during an electron transition in a hydrogen atom from n = 3 to n = 2, we can use the Rydberg formula for hydrogen emission. The formula is given by:

1 / λ = R * (1 / n_{1}^{2} - 1 / n_{2}^{2})

where λ is the wavelength, R is the Rydberg constant (1.097 x 107 m-1), n_{1} is the initial energy level, and n_{2} is the final energy level.

By substituting n_{1} = 2 and n_{2} = 3 into the equation, and solving for λ, we can find the wavelength of the emitted photon.

After calculating, we find that the wavelength is approximately 656.5 nm, which corresponds with the red line in the Balmer series of the hydrogen spectrum. Thus, the answer to the student's question is 657 nm (after rounding to the nearest nm).

Answer 2

657nm

Explanation

wavelength= 1/wavenumber

but wavenumber = RH×Z^2( (1/n1^2)-(1/n2^2))

where RH is the Rydberg constant = 1.0976×10^7

Z is the atomic number

n1 & n2 are orbitals

=(1.0976×10^7)×(1^2)×(1/4-1/9)

=1524444.444

but wavelength= 1/wavenumber

Therefore wavelength = 1/1524444.444

= 6.5597× 10^-7m

≈ 657nm