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

Explanation:

It is given that,

Momentum of the photon, $p=5.55\times 10^{-27}\ kg-m/s$

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

$\lambda=\dfrac{h}{p}$

h is the Planck's constant

$\lambda=\dfrac{6.67\times 10^{-34}\ Js}{5.55\times 10^{-27}\ kg-m/s}$

$\lambda=1.2\times 10^{-7}\ m$

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.