Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, illuminate a metal of work function 0.5 eV. The ratio of the maximum KE of the emitted electrons will be:

The correct option is 1:4.

To find the ratio of the maximum kinetic energies of the emitted electrons, we can use Einstein's photoelectric equation:
$$K_{\text{max}}=E-\Phi$$
where:

• $K_{\text{max}}$ is the maximum kinetic energy of the emitted photoelectrons,
• $E$ is the energy of the incident photons, and
• $\Phi$ is the work function of the metal.

We are given:

• Work function of the metal, $\Phi =0.5\text{eV}$
• Energy of the first light photons, $E_1=1\text{eV}$
• Energy of the second light photons, $E_2=2.5\text{eV}$

First, let's calculate the maximum kinetic energy ($K_1$) for the first frequency of light:
$$K_1=E_1-\Phi$$
$$K_1=1\text{eV}-0.5\text{eV}=0.5\text{eV}$$

Next, let's calculate the maximum kinetic energy ($K_2$) for the second frequency of light:
$$K_2=E_2-\Phi$$
$$K_2=2.5\text{eV}-0.5\text{eV}=2.0\text{eV}$$

Now, we find the ratio of these maximum kinetic energies:
$$\frac{K_1}{K_2}=\frac{0.5\text{eV}}{2.0\text{eV}}=\frac{1}{4}$$
Thus, the ratio of the maximum kinetic energies of the emitted electrons is 1:4.