The maximum velocity of the photoelectrons emitted from the surface is $$v$$ when light of frequency $$n$$ falls on a metal surface. If the incident frequency is increased to $$3n$$, the maximum velocity of the ejected photoelectrons will be:

We start with Einstein’s photo-electric equation, which relates the energy of an incident photon to the maximum kinetic energy of an emitted photoelectron:

$$h\nu = \phi + \dfrac{1}{2} m v_{\text{max}}^{\,2}$$

Here $$h$$ is Planck’s constant, $$\nu$$ (read “n”) is the frequency of the incident light, $$\phi$$ is the work function of the metal, $$m$$ is the mass of an electron and $$v_{\text{max}}$$ is the maximum speed of the ejected electrons.

First, for the given frequency $$\nu$$, the maximum velocity is given to be $$v$$. So we write

$$h\nu = \phi + \dfrac{1}{2} m v^{2}\;. \quad -(1)$$

Now the frequency of the incident light is increased to $$3\nu$$. Let the new maximum velocity be $$v'$$. Applying the same photo-electric equation for this second case gives

$$h(3\nu) = \phi + \dfrac{1}{2} m {v'}^{2}\;. \quad -(2)$$

We wish to compare $$v'$$ with $$v$$. To do so, we first isolate the kinetic-energy terms on the right in both equations and then subtract equation (1) from equation (2).

Subtracting (1) from (2) gives

$$h(3\nu) - h\nu = \left[\phi + \dfrac{1}{2} m {v'}^{2}\right] - \left[\phi + \dfrac{1}{2} m v^{2}\right].$$

Notice that $$\phi$$ cancels out:

$$2h\nu = \dfrac{1}{2} m {v'}^{2} - \dfrac{1}{2} m v^{2}.$$

Multiplying every term by $$2$$ to remove the half factors, we get

$$4h\nu = m{v'}^{2} - m v^{2}.$$

Dividing through by $$m$$ gives a direct relation between the squares of the velocities:

$${v'}^{2} - v^{2} = \dfrac{4h\nu}{m}\;. \quad -(3)$$

But equation (1) can be rearranged to express $$h\nu$$ in terms of known quantities:

From (1),

$$h\nu = \phi + \dfrac{1}{2} m v^{2}.$$

Substituting this value of $$h\nu$$ into equation (3) yields

$${v'}^{2} - v^{2} = \dfrac{4}{m}\!\left(\phi + \dfrac{1}{2} m v^{2}\right).$$

Distributing the fraction gives

$${v'}^{2} - v^{2} = \dfrac{4\phi}{m} + 2v^{2}.$$

Now we move $$v^{2}$$ across to find $$v'^{2}$$ explicitly:

$${v'}^{2} = v^{2} + 2v^{2} + \dfrac{4\phi}{m} = 3v^{2} + \dfrac{4\phi}{m}.$$

Taking the positive square root (because speed is positive) gives

$$v' = \sqrt{\,3v^{2} + \dfrac{4\phi}{m}\,}\;. \quad -(4)$$

The term $$\dfrac{4\phi}{m}$$ is definitely positive because both $$\phi$$ and $$m$$ are positive physical quantities. Therefore the radicand in (4) is greater than $$3v^{2}$$. Consequently,

$$v' > \sqrt{3v^{2}} = \sqrt{3}\,v.$$

Thus, when the incident frequency is tripled from $$\nu$$ to $$3\nu$$, the maximum velocity of the ejected photoelectrons becomes more than $$\sqrt{3}v$$.

Hence, the correct answer is Option A.