Question 110: Dual Nature of Matter and Radiation

Question

A metallic surface is illuminated with radiation of wavelength $$\lambda$$, the stopping potential is $$V_{0}$$. If the same surface is illuminated with radiation of wavelength $$2 \lambda$$. the stopping potential becomes $$\frac{V_{o}}{4}$$. The threshold wavelength for this metallic surface will be

Accepted Answer

The photoelectric effect occurs when light (or more generally, electromagnetic radiation) incident on a metallic surface causes the ejection of electrons from the surface. The energy of the incident photons must be greater than the work function of the metal (denoted by $$\phi_0$$) for the electrons to be ejected.
The energy of the ejected electrons can be expressed as the difference between the energy of the incident photons and the work function of the metal. The maximum kinetic energy of the ejected electrons is given by:
$$K_{max} = h
u - \phi_0$$
where $$h$$ is the Planck's constant, $$
u$$ is the frequency of the incident light, and $$\phi_0$$ is the work function of the metal.
We can also write the maximum kinetic energy in terms of the stopping potential ($$V_0$$) and the elementary charge ($$e$$) of an electron:
$$K_{max} = eV_0$$
Equating these two expressions for the maximum kinetic energy, we get:
$$eV_0 = h
u - \phi_0$$
We can express the frequency $$
u$$ in terms of the speed of light $$c$$ and the wavelength $$\lambda$$:
$$
u = \frac{c}{\lambda}$$
Substituting this expression for frequency in the equation, we get:
$$eV_0 = h\frac{c}{\lambda} - \phi_0$$
Now, we have two cases:
The stopping potential is $$V_0$$, and the wavelength is $$\lambda$$.
The stopping potential is $$\frac{V_0}{4}$$, and the wavelength is $$2\lambda$$.
For the first case, we use the photoelectric effect equation as is:
$$eV_0 = \frac{hc}{\lambda} - \phi_0$$
For the second case, we replace $$V_0$$ with $$\frac{V_0}{4}$$ and $$\lambda$$ with $$2\lambda$$:
$$\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$$
Now we have two equations:
(1) $$eV_0 = \frac{hc}{\lambda} - \phi_0$$
(2) $$\frac{eV_0}{4} = \frac{hc}{2\lambda} - \phi_0$$
We can rewrite equation (1) as:
$$\frac{eV_0}{4} = \frac{hc}{4\lambda} - \frac{\phi_0}{4}$$
Now we can equate the two expressions for $$\frac{eV_0}{4}$$:
$$\frac{hc}{4\lambda} - \frac{\phi_0}{4} = \frac{hc}{2\lambda} - \phi_0$$
Now we isolate the terms containing $$\phi_0$$:
$$\frac{3\phi_0}{4} = \frac{hc}{4\lambda} - \frac{hc}{2\lambda}$$
$$\frac{3\phi_0}{4} = \frac{hc}{4\lambda}$$
Now we can write the work function $$\phi_0$$ in terms of the threshold wavelength $$\lambda_0$$:
$$\phi_0 = \frac{hc}{\lambda_0}$$
Substituting the expression for the work function $$\phi_0$$ in terms of the threshold wavelength $$\lambda_0$$ into the previous equation, we get:
$$\frac{3}{4} \frac{hc}{\lambda_0} = \frac{hc}{4\lambda}$$
Now we can cancel out the common terms $$hc$$ from both sides:
$$\frac{3}{4\lambda_0} = \frac{1}{4\lambda}$$
Next, we can cross-multiply to solve for $$\lambda_0$$:
$$3\lambda = \lambda_0$$
Thus, the threshold wavelength for this metallic surface is $$3\lambda$$.

Answer

$$3 \lambda$$

Answer

$$4 \lambda$$

Answer

$$\frac{3}{2} \lambda$$

Answer

$$\frac{\lambda}{4}$$

Dual Nature of Matter and Radiation

A metallic surface is illuminated with radiation of wavelength $\lambda, the stopping potential is V_{0}. If the same surface is illuminated with radiation of wavelength 2 \lambda. the stopping potential becomes \frac{V_{o}}{4}$. The threshold wavelength for this metallic surface will be

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