The de-Broglie wavelength of a particle having kinetic energy $$E$$ is $$\lambda$$. How much extra energy must be given to this particle so that the de-Broglie wavelength reduces to 75% of the initial value?

We start with the well-known de-Broglie relation for any material particle, which states that the wavelength associated with a particle is given by

$$\lambda=\frac{h}{p},$$

where $$h$$ is Planck’s constant and $$p$$ is the linear momentum of the particle.

For a non-relativistic particle, the kinetic energy $$E$$ is related to the momentum by the formula

$$E=\frac{p^{2}}{2m},$$

where $$m$$ is the mass of the particle. Solving this equation for $$p$$ gives

$$p=\sqrt{2mE}.$$

Substituting this expression for $$p$$ back into the de-Broglie formula, we obtain

$$\lambda=\frac{h}{\sqrt{2mE}}.$$

From this last equation we notice an important proportionality:

$$\lambda\propto\frac{1}{\sqrt{E}}.$$

Now, let the initial kinetic energy be $$E$$ and the corresponding wavelength be $$\lambda$$. Suppose we supply extra energy so that the new kinetic energy becomes $$E'$$ and the new wavelength becomes $$\lambda'$$. According to the problem statement, the wavelength is reduced to 75 % of its initial value, so we have

$$\lambda' = 0.75\,\lambda=\frac{3}{4}\,\lambda.$$

Using the proportionality $$\lambda\propto\frac{1}{\sqrt{E}}$$ for both the initial and final states, we can write

$$\frac{\lambda'}{\lambda}=\sqrt{\frac{E}{E'}}.$$

Substituting $$\lambda'/\lambda = 3/4$$ into this equation gives

$$\frac{3}{4}=\sqrt{\frac{E}{E'}}.$$

Now we square both sides to remove the square root:

$$\left(\frac{3}{4}\right)^{2} = \frac{E}{E'}.$$

This simplifies to

$$\frac{9}{16} = \frac{E}{E'}.$$

Rearranging to solve for $$E'$$ gives

$$E' = \frac{16}{9}\,E.$$

The extra energy that must be supplied to the particle is the difference between the final and initial energies:

$$\text{Extra energy}=E' - E=\frac{16E}{9}-E=\frac{16E}{9}-\frac{9E}{9}=\frac{7E}{9}.$$

Hence, the correct answer is Option B.