What are the (a) energy, (b) magnitude of the momentum, and (c) wavelength of the photon emitted when a hydrogen atom undergoes a transition from a state with n = 3 to a state with n = 1 ?

Short Answer

Expert verified
  1. The required energy is 12.1 eV .
  2. The magnitude of the momentum is 6.46×10-27kg.m/s.
  3. The wavelength of the photon is 1.02×10-7m.

Step by step solution

01

Describe the energy of the photon:

The Energy of the photons is given by,

E=-13.6(1n32-1n12)

Here, n1 and n3 are the energy levels.

02

(a) Define the energy of the photon:

Write the equation of the energy of the photons as below.

E=-13.61n32-1n12 ….. (1)

Substitute 3 for n3, and 1 for n1in equation (1).

E=-13.6132-112=-13.619-1=-13.6-89=12.1eV

Therefore, the required energy is 12.1eV.

03

(b) Find the magnitude of the momentum:

The formula to calculate the magnitude of the momentum is given by,

p=Ec ….. (2)

Here, c is speed of the light.

Substitute 12.1eVfor E, and 3×108m/s for c in equation (2).

p=12.1eV3×108m/s=12.1×1.602×10-19J3×108m/s=19.38×10-19kgm2/s23×108m/s=6.46×10-27kg.m/s

Therefore, the magnitude of the momentum is 6.46×10-27kg.m/s.

04

(c) Determine the wavelength of the photon:

The formula to calculate the wavelength is given by,

λ=hp ….. (3)

Here, h is plank constant.

Substitute 6.626×10-34J.s for h, and 6.46×10-27kg.m/sfor p in equation (3).

λ=6.626×10-34J.s6.46×10-27kg·m/s=1.02×10-7m

Therefore, the wavelength of the photon is 1.02×10-7m.

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