Question

A homogeneous ball (mass=m) of ideal black material at rest is illuminated with a radiation having a set of photons (wavelength=λ), each with the same momentum and the same energy. The rate at which photons fall on the ball is n. The linear acceleration of the ball is:

Answer 1

Hint:Using the de-Broglie relation, this states that according to wave particle duality a photon which carries energy is the product of Planck constant h and frequency of wavelength \[\lambda \]. Now to find the acceleration we need the value of velocity which is
\[v=\dfrac{h}{m\lambda }\]
And linear momentum
\[p=mv\]
where m is the mass of the particle, v is the velocity of the particle, h is the Planck’s constant and \[\lambda \] are wavelengths.

Complete step by step solution:
Using the de-Broglie relation, when the photon falls on the ball, the linear momentum of the photon is given in terms of de-Broglie's wavelength with the de-Broglie’s wavelength as:
\[p\text{ }=\text{ }m\text{ }\times \text{ }v\]
With one photon, the wavelength of the lambda is:
\[\lambda =\dfrac{h}{p}\]
Hence, the velocity of the photon travelling from one point to another is given as:
\[v=\dfrac{h}{m\lambda }\]
Now placing the value of velocity and rate of fall of photon into the acceleration as
Acceleration = Rate of fall of photon × velocity of the photon. Placing the values in the formula, we get:
Acceleration = n × \[\dfrac{h}{m\lambda }\]

Therefore, the linear acceleration of the ball is \[\dfrac{nh}{m\lambda }\].

Note: The de Broglie wavelength is inversely proportional to the momentum. Hence, the acceleration is not dependent on the de Broglie wavelength. However, when the particle accelerates the velocity changes due to loss of energy due to radiation hence, the acceleration is dependent on wavelength and Planck's constant.