Animated Solution for Physics - Dual Nature of Matter and Radiation: In a photoelectric effect experiment, the threshold wavelength of light is $380\text{ nm}$. If the wavelength of incident light is $260\text{ nm}$, the maximum kinetic energy of emitted electrons will be Given, $E\text{ (in eV)}=\frac{1237}{\lambda \text{ (in nm)}}$

Imagine you are observing a fascinating quantum phenomenon: the photoelectric effect. We have a specific metal surface, and we are told its threshold wavelength (${\lambda }_0$) is $380\text{ nm}$. This threshold is a critical boundary. It tells us the absolute maximum wavelength a photon can have and still possess enough energy to knock an electron free from the metal's grip.

Now, we shine an incident light onto this surface with a wavelength ($\lambda$) of $260\text{ nm}$. Because $260\text{ nm}$ is shorter than $380\text{ nm}$, we know immediately that our incident photons pack a stronger energetic punch than the minimum required. Electrons will definitely be ejected!

To find out exactly how much kinetic energy these liberated electrons will carry, we turn to Albert Einstein's elegant photoelectric equation. The principle is a simple conservation of energy:

Here, $E_{\text{incident}}$ is the energy of the incoming photon, and $\Phi$ is the work function (the minimum energy required to break the electron free). The leftover energy becomes the maximum kinetic energy ($K{E}_{\text{max}}$) of the ejected electron.

The problem generously provides a shortcut formula to calculate energy directly in electron-volts (eV) when the wavelength is in nanometers:

Let's substitute this handy formula into Einstein's equation for both the incident photon and the work function (which corresponds to the threshold wavelength):

Now, we plug in our specific values: $\lambda =260\text{ nm}$ and ${\lambda }_0=380\text{ nm}$.

To avoid a messy calculation, let's factor out the common numerator, $1237$. This is a crucial step to prevent silly arithmetic mistakes under exam pressure.

Now, we simply resolve the fraction inside the parentheses by finding a common denominator:

By multiplying and dividing these final numbers, we arrive at our destination:

The maximum kinetic energy of the emitted electrons is approximately $1.5\text{ eV}$, which perfectly matches option (c).

As a final thought experiment, consider what would happen if our incident light had a wavelength of $400\text{ nm}$. Because $400\text{ nm}$ is greater than the threshold of $380\text{ nm}$, the incident photons would lack the energy to overcome the work function. In that scenario, absolutely no photoelectric emission would occur, regardless of how bright the light was!