Question

When a photon is emitted by a hydrogen atom, the photon carries a momentum with it. (a) Calculate the momentum carried by the photon when a hydrogen atom emits light of wavelength 656.3 nm. (b) With what speed does the atom recoil during this transition? Take the mass of the hydrogen atom = \[1.67 \times {10^{ - 27}}\]kg. (c) Find the kinetic energy of recoil of the atom

Answer 1

Hint:The subatomic particle like a photon shows duality nature while in motion. It behaves as the wave and particle. To find the momentum we can use the De-Broglie formula as well as the classical formula for momentum.

Formula used:
\[P = \dfrac{E}{c}\]
Where P is the momentum of a photon with energy E and c is the speed of light.
\[P = mv\]
where P is the momentum of the particle with mass m and speed v.
\[K = \dfrac{1}{2}m{v^2}\]
where K is the kinetic energy of the particle with mass m and speed v.

Complete step by step solution:
(a) The wavelength of the light emitted by light is given as 656.3 nm.
\[\lambda = 656.3\,nm\]
We need to find the momentum carried by the photon of light emitted by the hydrogen atom. Using the formula of momentum,
\[P = \dfrac{E}{c}\]
\[\Rightarrow P = \dfrac{{\left( {\dfrac{{hc}}{\lambda }} \right)}}{c} \\ \]
\[\Rightarrow P = \dfrac{h}{\lambda } \\ \]
Putting the values, we get
\[P = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{656.3 \times {{10}^{ - 9}}}}kgm/s \\ \]
\[\therefore P = 1.01 \times {10^{ - 27}}\,kgm/s\]

Therefore, the momentum carried by the photon emitted by the hydrogen atom is \[1.01 \times {10^{ - 27}}kgm/s\].

(b) We got the momentum of the photon in the first part of the question. We have asked to find the speed of the photon with which it collides. As we know that the magnitude of momentum is the product of the mass of the particle and the speed of the particle. So, using the formula of momentum,
\[P = mv\]
\[\Rightarrow v = \dfrac{P}{m}\]
The mass of the photon is given as,
\[m = 1.67 \times {10^{ - 27}}kg\]
Putting the values, we get
\[v = \dfrac{{1.01 \times {{10}^{ - 27}}}}{{1.67 \times {{10}^{ - 27}}}}m/s \\ \]
\[\therefore v = 0.6\,m/s\]

Therefore, the speed of the photon is 0.6 m/s.

(c) We got the speed of the photon. We need to find the kinetic energy of the photon.
Using the formula of kinetic energy, we get
\[K = \dfrac{1}{2}m{v^2} \\ \]
\[\Rightarrow K = \dfrac{1}{2} \times \left( {1.67 \times {{10}^{ - 27}}} \right) \times \left( {{{0.6}^2}} \right)J \\ \]
\[\therefore K = 3.01 \times {10^{ - 28}}J\]

Therefore, the kinetic energy of the emitted photon is \[3.01 \times {10^{ - 28}}J\].

Note: If the speed of photons is comparable to the speed of light then we can’t use the non-relativistic formula to find the kinetic energy and the momentum. In that case we need to consider relativity.