Two stream of photons, possessing energies equal to twice and ten times the work function of metal a

Q: Why do we subtract the work function from the incident energy?

The work function is the minimum energy required just to free the electron from the metal's surface. Any leftover energy from the incident photon becomes the kinetic energy of the ejected electron.

Q: Why does the mass of the electron not affect the final ratio?

Since we are comparing the velocities of electrons in both cases, the mass $m$ is constant and cancels out when we take the ratio of their kinetic energies.

Q: What if the incident energy was less than the work function?

If the incident energy is less than the work function, no electrons would be emitted, regardless of the intensity of the light. This is a core principle of the photoelectric effect!

Q: Why do we take the square root of the kinetic energy ratio?

Kinetic energy is given by $\frac{1}{2}mv^2$. Therefore, the ratio of kinetic energies is equal to the ratio of the squares of the velocities. To find the ratio of the velocities themselves, we must take the square root.

Q: Can the ratio $x:y$ be something like $2:6$?

While mathematically equivalent, ratios in such physics problems are conventionally expressed in their simplest integer form unless specified otherwise. Hence, $1:3$ gives $x=1$.

Question

Two stream of photons, possessing energies equal to twice and ten times the work function of metal are incident on the metal surface successively. The value of ratio of maximum velocities of the photoelectrons emitted in the two respective cases is $x : y$. The value of $x$ is .............. .

Sigma Percentile · Accepted Answer

### The Magic of the Photoelectric Effect

Imagine you are observing a pristine metal surface at the atomic level. Suddenly, a beam of light strikes it, and like magic, electrons are ejected into the void! This beautiful phenomenon is known as the **photoelectric effect**.

To understand this, we rely on Einstein's elegant photoelectric equation. It states that the energy of the incoming photon is split into two parts. First, it pays the "toll tax" required to free the electron, known as the **work function** ($\phi_0$). Whatever energy is left over becomes the **maximum kinetic energy** ($K_{max}$) of the escaping electron.

Mathematically, this is written as:

$$E = \phi_0 + K_{max}$$

### Setting Up the First Scenario

In our problem, we are given two distinct scenarios. Let's tackle them one by one.

In the first case, the incident photon arrives with an energy equal to twice the work function, so $E_1 = 2\phi_0$.

We substitute this into our master equation:

$$2\phi_0 = \phi_0 + \frac{1}{2}mv_1^2$$

By simply subtracting the work function from both sides, we find the kinetic energy of the electron in the first case:

$$\frac{1}{2}mv_1^2 = \phi_0$$

### Unleashing the Second Photon

Now, let's move on to the second case. Here, the incident photon is an absolute powerhouse, carrying an energy ten times the work function! So, $E_2 = 10\phi_0$.

We set up the equation for this high-energy scenario:

$$10\phi_0 = \phi_0 + \frac{1}{2}mv_2^2$$

Again, we subtract the work function to find the kinetic energy of this much faster electron:

$$\frac{1}{2}mv_2^2 = 9\phi_0$$

### The Grand Comparison

We now have the kinetic energies for both cases. The question asks for the ratio of their maximum velocities. To find this, we divide the kinetic energy of the first case by that of the second case.

$$\frac{\frac{1}{2}mv_1^2}{\frac{1}{2}mv_2^2} = \frac{\phi_0}{9\phi_0}$$

Notice how beautifully the mass ($m$) and the $\frac{1}{2}$ cancel out on the left side, while the work function ($\phi_0$) cancels out on the right side. We are left with a clean ratio of the squares of the velocities:

$$\frac{v_1^2}{v_2^2} = \frac{1}{9}$$

### The Final Reveal

Remember, kinetic energy is proportional to the square of the velocity. To find the ratio of the velocities themselves, we simply take the square root of both sides.

$$\frac{v_1}{v_2} = \sqrt{\frac{1}{9}} = \frac{1}{3}$$

The problem states that this ratio is $x : y$, which means $\frac{x}{y} = \frac{1}{3}$.

By direct comparison, we can clearly see that $x = 1$ and $y = 3$. Therefore, the value of $x$ is **1**. And there we have it, a flawless victory over the photoelectric effect!

Enter Numerical Value:

Visualized Solution

$P h o t oe l ec t r i c E f f ec t$

$C a se 1 : F i r s t P h o t o n$

$C a se 1 : K in e t i c E n er g y$

$C a se 2 : S eco n d P h o t o n$

$C a se 2 : K in e t i c E n er g y$

$R a t i o o f K in e t i c E n er g i es$

$R a t i o o f V e l oc i t i es$

$F ina l A n s w er$

The Sigma Insight: Photoelectric Effect

The Magic of the Photoelectric Effect

Setting Up the First Scenario

Unleashing the Second Photon

The Grand Comparison

The Final Reveal

Similar Questions

The surface of a metal is illuminated alternately with photons of energies $E_{1} = 4 eV$ and $E_{2} = 2.5 eV$ respectively. The ratio of maximum speeds of the photoelectrons emitted in the two cases is 2. The work function of the metal (in eV) is ......... .

When the wavelength of radiation falling on a metal is changed from $500 nm$ to $200 nm$ , the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Photoelectric Effect

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: First Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Second Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: Kinetic Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Kinetic Energies

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Ratio of Velocities

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Answer