Question 357: Dual Nature of Matter and Radiation

Question

When a light of wavelength '$$\lambda$$' falls on the emitter of a photocells, maximum speed of emitted photoelectrons is '$$\mathrm{V}$$'. If the incident wavelength is changed to $$\frac{2 \lambda}{3}$$, maximum speed of emitted photoelectrons will be

Accepted Answer

Here's a detailed explanation of the solution:
The photoelectric effect is governed by Einstein's photoelectric equation:
$$h
u = \phi + \frac{1}{2}mv^2$$
where:
$$h$$ is Planck's constant
$$ÃÂ½$$ 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 $$c = ÃÂ½ÃÂ»$$ (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 $$v(1.5)^{\frac{1}{2}}$$

Answer

less than $$\mathrm{V}(1.5)^{1 / 2}$$

Answer

$$\sqrt{\mathrm{V}}$$

Answer

grater than $$\mathrm{V}(1.5)^{\frac{1}{2}}$$

Answer

$$\mathrm{V}$$

Dual Nature of Matter and Radiation

When a light of wavelength '$\lambda' falls on the emitter of a photocells, maximum speed of emitted photoelectrons is '\mathrm{V}'. If the incident wavelength is changed to \frac{2 \lambda}{3}$, maximum speed of emitted photoelectrons will be

Choose the correct answer: