$$h$$ is Planck's constant

$$ν$$ is the frequency of the incident light

$$\phi$$ is the work function of the metal (the minimum energy required to eject an electron)

$$m$$ is the mass of the electron

$$v$$ is the maximum speed of the emitted photoelectron

We can rewrite the equation in terms of wavelength using the relationship $$c = νλ$$ (where $$c$$ is the speed of light):

$$hc/\lambda = \phi + \frac{1}{2}mv^2$$

Let's analyze the scenario:

Initially, with wavelength $$\lambda$$, the maximum speed is $$v$$. So we have:

$$hc/\lambda = \phi + \frac{1}{2}mv^2$$

Now, the wavelength is changed to $$\frac{2 \lambda}{3}$$. Let's call the new maximum speed $$v'$$. The equation becomes:

$$hc/(\frac{2 \lambda}{3}) = \phi + \frac{1}{2}mv'^2$$

Simplifying the second equation:

$$ \frac{3hc}{2 \lambda} = \phi + \frac{1}{2}mv'^2$$

Notice that the work function ($$\phi$$) remains the same for the same metal. Subtracting the first equation from the second equation, we get:

$$\frac{3hc}{2 \lambda} - \frac{hc}{\lambda} = \frac{1}{2}mv'^2 - \frac{1}{2}mv^2$$

Simplifying further:

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

We can now see that the left-hand side of the equation has increased by a factor of 1.5. Therefore, the right-hand side must also increase by a factor of 1.5. This means:

$$v'^2 - v^2 = 1.5v^2$$

Solving for $$v'$$:

$$v'^2 = 2.5v^2$$

$$v' = \sqrt{2.5}v = \sqrt{1.5^2}v = 1.5v$$

Therefore, the maximum speed of emitted photoelectrons will be greater than $$v(1.5)^{\frac{1}{2}}$$.

The correct answer is Option C: greater than $$v(1.5)^{\frac{1}{2}}$$

Dual Nature Of Radiation

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