Question

A photon has a momentum of 5.55 x 10-27 kg-m/s. (a) What is the photon's wavelength? nm (b) To what part of the electromagnetic spectrum does this wavelength correspond? the tolerance is +/-2%

Answer 1

Step-by-step explanation:

It is given that,

Momentum of the photon,
`p=5.55* 10^(-27)\ kg-m/s`

(a) We need to find the wavelength of this photon. It can be calculated using the concept of De-broglie wavelength.

`\lambda=(h)/(p)`

h is the Planck's constant

`\lambda=(6.67* 10^(-34)\ Js)/(5.55* 10^(-27)\ kg-m/s)`

`\lambda=1.2* 10^(-7)\ m`

or

`\lambda=120\ nm`

(b) The wavelength lies in the group of ultraviolet rays. The wavelength of UV rays lies in between 400 nm to 10 nm.