The surface of a metal is illuminated alternately with photons of energies E1 = 4 eV and E2 = 2.5 eV

Q: Why do we use the 'maximum' kinetic energy in the formula?

Electrons inside the metal can lose some energy due to internal collisions before they reach the surface and escape. The equation $K_{\max} = E - \phi$ assumes the best-case scenario where an electron at the very surface absorbs the photon and escapes without any collision losses.

Q: Does the mass of the electron affect the final answer?

No, it doesn't. When we set up the ratio of the two kinetic energies, the $\frac{1}{2}m$ term appears in both the numerator and the denominator, perfectly canceling out. The problem relies purely on the energy relationships.

Q: Why did we square the velocity ratio?

Kinetic energy is given by $\frac{1}{2}mv^2$. Therefore, the ratio of kinetic energies is proportional to the square of the velocities: $\frac{K_1}{K_2} = \frac{v_1^2}{v_2^2} = \left(\frac{v_1}{v_2}\right)^2$. Since $v_1/v_2 = 2$, the kinetic energy ratio is $2^2 = 4$.

Q: What would happen if we used a photon with an energy of 1.5 eV on this metal?

Since we calculated the work function to be 2 eV, a 1.5 eV photon does not have enough energy to overcome the binding energy of the electron. No photoelectric emission would occur.

Q: Can the work function ever be negative?

No. The work function represents the minimum energy required to pull an electron out of the metal's attractive potential well. It is always a positive value for any material.

Question

The surface of a metal is illuminated alternately with photons of energies $E_1 = 4	ext{ eV}$ and $E_2 = 2.5	ext{ eV}$ 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 ......... .

Sigma Percentile · Accepted Answer

### Unveiling the Work Function: A Tale of Two Photons

Imagine you are standing in a microscopic shooting gallery. Your targets are electrons bound to a metal surface, and your ammunition consists of photons—tiny packets of light energy. When a photon strikes the metal with enough energy, it can knock an electron completely out of its atomic bounds. This beautiful phenomenon is known as the **photoelectric effect**, and it serves as the foundation for this intriguing problem.

### The Master Equation

To navigate this quantum landscape, we rely on Albert Einstein's elegant photoelectric equation. It states that the maximum kinetic energy ($K_{\max}$) of an ejected electron is equal to the energy of the incident photon ($E$) minus the energy required to break the electron free, known as the work function ($\phi$).

$$K_{\max} = E - \phi$$

This equation is essentially a statement of the conservation of energy. The photon gives all its energy to the electron. The electron pays the "toll" (the work function) to escape the metal, and whatever energy is left over becomes its kinetic energy.

### Analyzing the Two Cases

In our problem, we are conducting two separate experiments on the same metal surface.

**Case 1:** We fire a high-energy photon with $E_1 = 4	ext{ eV}$. The electron escapes with a maximum velocity $v_1$. We can write the energy balance as:

$$\frac{1}{2}mv_1^2 = 4 - \phi$$

**Case 2:** We switch to a lower-energy photon with $E_2 = 2.5	ext{ eV}$. The electron now escapes with a slower maximum velocity $v_2$. The energy balance becomes:

$$\frac{1}{2}mv_2^2 = 2.5 - \phi$$

### The Crucial Clue

We are given a vital piece of information: the ratio of the maximum speeds in the two cases is 2. This means the first electron is moving exactly twice as fast as the second one ($v_1 / v_2 = 2$).

To utilize this clue, let's divide our first kinetic energy equation by the second one. Notice how the $\frac{1}{2}m$ terms elegantly cancel out, leaving us with a ratio of squared velocities:

$$\frac{v_1^2}{v_2^2} = \frac{4 - \phi}{2.5 - \phi}$$

Since $v_1 / v_2 = 2$, squaring this ratio gives us 4. Substituting this into our equation yields:

$$4 = \frac{4 - \phi}{2.5 - \phi}$$

### Final Calculation

Now, we are left with a straightforward algebraic equation. Let's cross-multiply carefully to avoid any silly mistakes:

$$4(2.5 - \phi) = 4 - \phi$$

Expanding the left side:

$$10 - 4\phi = 4 - \phi$$

Rearranging the terms to isolate $\phi$:

$$10 - 4 = 4\phi - \phi$$

$$6 = 3\phi$$

Dividing by 3, we arrive at our final answer:

$$\phi = 2	ext{ eV}$$

The work function of this mysterious metal is exactly **$2	ext{ eV}$**. This means any photon with an energy less than $2	ext{ eV}$ will simply bounce off or be absorbed as heat, completely failing to eject a single electron. The quantum toll must always be paid in full!

Enter Numerical Value:

Visualized Solution

Visualizing the Photoelectric Effect

Einstein's Photoelectric Equation

Setting up Case 1

Setting up Case 2

Using the Speed Ratio

Dividing the Equations

Substituting the Ratio

Cross-Multiplication

Final Calculation

The Way Forward

The Sigma Insight: Photoelectric Effect

Unveiling the Work Function

The Master Equation

Analyzing the Two Cases

The Crucial Clue

Final Calculation

Similar Questions

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