found by the equation:

$KE = hf - \phi$

where $hf$ is the energy of the incident photon, and $\phi$ is the work function of the metal.

Since we know the energies of the photons are in multiples of the work function ($\phi$): the first photon has an energy of $2\phi$ and the second has an energy of $5\phi$, we can substitute these into the equation above to find the kinetic energies of the emitted photoelectrons in each case.

For the first photon:

$KE_1 = 2\phi - \phi = \phi$

For the second photon:

$KE_2 = 5\phi - \phi = 4\phi$

The kinetic energy of a photoelectron is also given by the equation:

$KE = \frac{1}{2}mv^2$

where $m$ is the mass of the electron, and $v$ is the velocity of the photoelectron. Therefore, we can equate the expressions for kinetic energy derived from the photoelectric effect to this kinetic energy formula to compare the velocities.

For the first photon:

$\phi = \frac{1}{2}mv_1^2$

For the second photon:

$4\phi = \frac{1}{2}mv_2^2$

To find the ratio $v_1:v_2$, we solve for $v_1$ and $v_2$ and take the ratio. From $KE = \frac{1}{2}mv^2$, we get $v = \sqrt{\frac{2KE}{m}}$. Therefore, the velocities are:

$v_1 = \sqrt{\frac{2\phi}{m}}$

$v_2 = \sqrt{\frac{2\cdot 4\phi}{m}} = \sqrt{4}\sqrt{\frac{2\phi}{m}} = 2\sqrt{\frac{2\phi}{m}}$

So, the ratio of $v_1$ to $v_2$ is $\frac{\sqrt{\frac{2\phi}{m}}}{2\sqrt{\frac{2\phi}{m}}} = \frac{1}{2}$.

Hence, the correct option is Option B: 1 : 2.

Dual Nature Of Radiation

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