Animated Solution for Physics - Dual Nature of Matter and Radiation: When the wavelength of radiation falling on a metal is changed from $500\text{ nm}$ to $200\text{ nm}$, the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

Imagine you are standing in front of a mysterious metal vault. You have a flashlight, and you start shining it on the metal. Suddenly, tiny invisible particles—electrons—start popping out of the surface! This isn't magic; this is the Photoelectric Effect, a phenomenon that completely revolutionized our understanding of light and won Albert Einstein his Nobel Prize.

In this problem, we are playing with the wavelength of the light we shine on the metal. We are told that when we change the wavelength from $500\text{ nm}$ to $200\text{ nm}$, the maximum kinetic energy of the ejected electrons becomes three times larger. Our mission? To find the work function (${\varphi }_0$) of the metal, which is essentially the "toll fee" an electron must pay to escape the metal's surface.

To solve this, we need Einstein's famous photoelectric equation:

This equation is beautifully simple. It says that the maximum kinetic energy ($K_{\text{max}}$) of an escaping electron is equal to the total energy of the incoming photon ($\frac{hc}{\lambda }$) minus the energy required to break free from the metal (${\varphi }_0$).

To make our calculations lightning fast, we use a brilliant shortcut for the product of Planck's constant ($h$) and the speed of light ($c$):

Let's break down the problem into the two cases provided.

Case 1: The incident wavelength is ${\lambda }_1=500\text{ nm}$. Let's assume the maximum kinetic energy is $K$. Plugging this into our master equation:

Case 2: The incident wavelength is reduced to ${\lambda }_2=200\text{ nm}$. Because the wavelength is shorter, the photons pack a much bigger punch! The problem states the kinetic energy becomes three times larger, so it is now $3K$:

Now we have a neat system of two linear equations. We want to find ${\varphi }_0$, so let's eliminate $K$. We can do this by substituting the expression for $K$ from the first equation directly into the second equation:

Let's carefully expand the bracket on the left side. Don't rush through this; watch out for the minus sign!

Next, we group the terms containing our unknown, ${\varphi }_0$, on one side, and the constant fractions on the other side. Moving $-3{\varphi }_0$ to the right side gives us $2{\varphi }_0$:

To avoid messy decimal arithmetic, let's factor out the common numerator, $1240$:

Finding a common denominator for the fractions inside the bracket (which is $1000$):

Finally, divide by $2$ to isolate the work function:

Looking at our options, $0.61\text{ eV}$ is the closest value (the slight difference arises because $hc$ is more precisely $1242\text{ eV}\cdot \text{nm}$, but $1240$ is the standard JEE approximation). Thus, the correct choice is (d).

Isn't it amazing how a few lines of algebra can reveal the hidden quantum properties of a metal?