Radiation of wavelength $$\lambda$$ is incident on a photocell. The fastest emitted photoelectron has a speed v. If the wavelength is changed to $$\frac{3\lambda}{4}$$, the speed of the fastest emitted photoelectron will be:

We start with the Einstein photo-electric equation, which gives the maximum (i.e., fastest) kinetic energy of the emitted photo-electrons.

Einstein’s equation is stated as

$$\frac12\,m\,v_{\text{max}}^{\,2}=h\nu-\phi,$$

where $$m$$ is the electron mass, $$v_{\text{max}}$$ is the speed of the fastest photo-electron, $$h$$ is Planck’s constant, $$\nu$$ is the frequency of the incident light, and $$\phi$$ is the work function of the metal surface.

For radiation of wavelength $$\lambda$$ we write the frequency as $$\nu=\dfrac{c}{\lambda},$$ so the above formula becomes

$$\frac12\,m\,v^{2}=h\dfrac{c}{\lambda}-\phi\quad\quad (1)$$

where $$v$$ is the given speed of the fastest electron for this wavelength.

Now the wavelength is changed to $$\dfrac{3\lambda}{4}.$$ Its frequency is therefore

$$\nu'=\dfrac{c}{\,3\lambda/4\,}=\dfrac{4c}{3\lambda}.$$

Using Einstein’s equation again we have for the new maximum speed $$v'$$

$$\frac12\,m\,v'^{2}=h\nu'-\phi=h\dfrac{4c}{3\lambda}-\phi\quad\quad (2)$$

We shall eliminate the work function $$\phi$$ between (1) and (2) so that everything is expressed in terms of the known speed $$v.$$(This is useful because $$\phi$$ is not numerically given.)

From equation (1) we isolate $$\phi$$:

$$\phi=h\dfrac{c}{\lambda}-\frac12\,m\,v^{2}\quad\quad (3)$$

Substituting (3) into the right-hand side of equation (2) gives

$$\frac12\,m\,v'^{2}=h\dfrac{4c}{3\lambda}-\Bigl[h\dfrac{c}{\lambda}-\frac12\,m\,v^{2}\Bigr].$$

We now remove the square brackets and gather like terms:

$$\frac12\,m\,v'^{2}=h\dfrac{4c}{3\lambda}-h\dfrac{c}{\lambda}+\frac12\,m\,v^{2}.$$

Because $$\dfrac{4}{3}-1=\dfrac{1}{3},$$ the first two terms combine neatly:

$$\frac12\,m\,v'^{2}=h\dfrac{c}{3\lambda}+\frac12\,m\,v^{2}.$$

To make comparison easier, multiply both sides by $$\dfrac{2}{m}$$:

$$v'^{2}=\dfrac{2h\,c}{3m\lambda}+v^{2}.$$

Thus the increase in the square of the speed is

$$v'^{2}-v^{2}=\dfrac{2h\,c}{3m\lambda}\quad\quad (4)$$

Now we use equation (1) again to express $$\dfrac{2h\,c}{m\lambda}$$ in terms of $$v^{2}$$ and $$\phi.$$ From (1) we have

$$\frac12\,m\,v^{2}=h\dfrac{c}{\lambda}-\phi \;\Longrightarrow\; \dfrac{2h\,c}{m\lambda}=v^{2}+\dfrac{2\phi}{m}.$$

Dividing by 3 gives

$$\dfrac{2h\,c}{3m\lambda}=\dfrac{1}{3}\Bigl(v^{2}+\dfrac{2\phi}{m}\Bigr).$$

Substituting this back into (4) we get

$$v'^{2}=v^{2}+\dfrac{1}{3}\Bigl(v^{2}+\dfrac{2\phi}{m}\Bigr) =\dfrac{4}{3}\,v^{2}+\dfrac{2\phi}{3m}.$$

Because the work function $$\phi$$ is always positive, the second term $$\dfrac{2\phi}{3m}$$ is also positive. Consequently

$$v'^{2} > \dfrac{4}{3}\,v^{2} \;\Longrightarrow\; v' > v\sqrt{\dfrac{4}{3}}.$$

Therefore the new fastest electron is emitted with a speed greater than $$v\bigl(\dfrac{4}{3}\bigr)^{1/2}$$ and not equal to or less than this value.

Hence, the correct answer is Option C.