When velocity of a particle is reduced by$2/3$. The wavelength associated with it.

Detailed solution :

The wavelength associated with a particle is related to its momentum, which is given by the de Broglie equation:

$\lambda = h/p$

where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle. The momentum of a particle is related to its velocity (v) and mass (m) by the formula:

$p = mv$

Therefore, if the velocity of a particle is reduced by 2/3, its momentum will becomes 1/3rd of the initial value. This means that the wavelength associated with the particle will increase by a factor of 3, according to the de Broglie equation:

$$\begin{gathered} \lambda \text{'} = h/(1/3)mv = 3h/mv \\ \Rightarrow \lambda \text{'} = 3\lambda \end{gathered}$$

So the new wavelength associated with the particle will be 3 times the original wavelength.

Hence, the correct option is C.