If $$\lambda_0$$ and $$\lambda$$ be threshold wavelength and wavelength of incident light, the velocity of photoelectron ejected from the metal surface is:

We start with the photoelectric effect equation. The maximum kinetic energy of the ejected photoelectron is given by the difference between the energy of the incident photon and the work function of the metal. The work function is the minimum energy required to eject an electron, which corresponds to the threshold wavelength λ₀.

The energy of the incident photon with wavelength λ is $$ E = \frac{hc}{\lambda} $$, where h is Planck's constant and c is the speed of light. The work function φ is $$ \phi = \frac{hc}{\lambda_0} $$. Therefore, the kinetic energy (K.E.) is:

$$ K.E. = \frac{hc}{\lambda} - \frac{hc}{\lambda_0} $$

Simplifying this expression:

$$ K.E. = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_0} \right) $$

Note that $$ \frac{1}{\lambda} - \frac{1}{\lambda_0} = \frac{\lambda_0 - \lambda}{\lambda \lambda_0} $$, so we can write:

$$ K.E. = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

This kinetic energy is also equal to $$ \frac{1}{2} m v^2 $$, where m is the mass of the electron and v is its velocity. Setting them equal:

$$ \frac{1}{2} m v^2 = hc \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

Now, solve for v²:

$$ v^2 = \frac{2 hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) $$

Taking the square root of both sides to find v:

$$ v = \sqrt{ \frac{2 hc}{m} \left( \frac{\lambda_0 - \lambda}{\lambda \lambda_0} \right) } $$

Comparing this with the given options, we see that it matches option C:

$$ \sqrt{\frac{2hc}{m}\left(\frac{\lambda_0 - \lambda}{\lambda \lambda_0}\right)} $$

Option A lacks the factor c and has incorrect units. Option B has an extra (λ₀ - λ) without the denominator, leading to incorrect units. Option D has the reciprocal terms reversed and lacks c, making it incorrect both dimensionally and physically.

Hence, the correct answer is Option C.