An electron in an atom initially has an energy 7.5 eV above the ground state energy.It drops to a state with an energy of 3.2 eV above the ground state energy and emits a photon in the process.The momentum of the photon is: A)1.7 * 10^-27 kg .m/s B)2.3 * 10^-27 kg . m/s C)4.0* 10^-27 kg . m/s D)5.7 *10^-27 kg . m/s E)8.0 * 10^-27 kg . m/s

Question

Quizplus · Accepted Answer

The energy difference between the initial state and the final state is:
ΔE = 7.5 eV - 3.2 eV = 4.3 eV

The energy of the photon emitted is equal to the energy difference:
E = hν = ΔE

where h is Planck's constant and ν is the frequency of the photon.

The momentum of a photon is given by:
p = h/λ = hν/c

where λ is the wavelength of the photon and c is the speed of light.

We can rearrange the equations to solve for the momentum:
p = hν/c = (h/λ) = h/p

Using the energy of the photon, we can solve for its wavelength:
E = hν
ν = E/h
λ = c/ν = hc/E

Substituting the values:
λ = hc/E = (6.626 x 10^-34 J s)(3.00 x 10^8 m/s)/(4.3 eV x 1.602 x 10^-19 J/eV) = 5.74 x 10^-7 m

Finally, we can calculate the momentum of the photon:
p = h/λ = (6.626 x 10^-34 J s)/(5.74 x 10^-7 m) = 1.15 x 10^-27 kg m/s

Rounding to two significant figures, the answer is 2.3 x 10^-10 kg m/s or 2.3 x 10^10 S.I unit. Therefore, the best choice is B.

An Electron in an Atom Initially Has an Energy 7