In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R

Q: Why do metals $P$ and $Q$ have the same incident energy $E_1$?

The problem explicitly states that the "same source of monochromatic light is used for metals $P$ and $Q$." Monochromatic light consists of photons that all have the exact same energy.

Q: Why did we multiply the entire $(E_1 - 4.5)$ expression by 2?

The given relation is $E_P = 2E_Q$. Since $E_Q$ is the entire expression $(E_1 - 4.5)$, you must substitute it in parentheses and distribute the 2 to both terms to maintain algebraic equality.

Q: How did we know $E_R$ is equal to $E_Q$?

The problem provides the continuous equality $E_P = 2E_Q = 2E_R$. If you look at the second part of that equation, $2E_Q = 2E_R$, you can simply divide both sides by 2 to get $E_Q = E_R$.

Q: What would happen if $E_1$ was calculated to be less than $4.5 \text{ eV}$?

If $E_1$ was less than the work function of metal $Q$ ($4.5 \text{ eV}$), the kinetic energy $E_Q$ would be negative. Physically, this means the incident photons wouldn't have enough energy to eject electrons from metal $Q$, and no photoemission would occur.

Q: Why don't we need to convert electron-volts (eV) to Joules?

Since all the given values (work functions and kinetic energy relations) are in electron-volts, and the final answer is requested in electron-volts, we can keep all our calculations in eV. It acts as a perfectly valid, self-consistent unit of energy for atomic physics.

Question

In a photoemission experiment, the maximum kinetic energies of photoelectrons from metals P, Q and R are $E_P$, $E_Q$ and $E_R$, respectively, and they are related by $E_P = 2E_Q = 2E_R$. In this experiment, the same source of monochromatic light is used for metals P and Q while a different source of monochromatic light is used for the metal R. The work functions for metals P, Q and R are 4.0 eV, 4.5 eV and 5.5 eV, respectively. The energy of the incident photon used for metal R, in eV, is _________.

Sigma Percentile · Accepted Answer

The photoelectric effect is one of the most profound phenomena in modern physics, serving as the definitive proof of the particle nature of light. When a photon strikes a metal surface, it transfers its energy to an electron. If this energy is sufficient to overcome the metal's binding energy—known as the work function—the electron is liberated. Any leftover energy manifests as the electron's kinetic energy. This elegant conservation of energy is encapsulated in Einstein's Photoelectric Equation.

In this problem, we are presented with a fascinating multi-scenario setup. We have three different metals: $P$, $Q$, and $R$. The problem states that metals $P$ and $Q$ are illuminated by the **same** monochromatic light source. This is a critical piece of information. 'Monochromatic' means light of a single wavelength, which implies that every single photon in that beam carries the exact same amount of energy. Let's denote the energy of these incident photons as $E_1$.

On the other hand, metal $R$ is illuminated by a **different** monochromatic light source. We will call the energy of its photons $E_2$. Our ultimate goal is to find the value of $E_2$.

### The Master Equation

Before we dive into the algebra, let's write down our primary tool. Einstein's Photoelectric Equation is given by:

$$K_{\max} = E_{	ext{incident}} - \phi$$

where $K_{\max}$ is the maximum kinetic energy of the emitted photoelectron, $E_{	ext{incident}}$ is the energy of the incoming photon, and $\phi$ is the work function of the specific metal.

### Setting Up the Algebra

Let's apply this master equation to our first two metals, $P$ and $Q$. We are given their respective work functions: $\phi_P = 4.0 	ext{ eV}$ and $\phi_Q = 4.5 	ext{ eV}$.

For metal $P$, the maximum kinetic energy $E_P$ can be written as:

$$E_P = E_1 - 4.0$$

For metal $Q$, the maximum kinetic energy $E_Q$ is:

$$E_Q = E_1 - 4.5$$

At this stage, we have two equations but three unknowns ($E_P$, $E_Q$, and $E_1$). We need a bridge to connect them.

### The Crucial Link

The problem provides a beautiful mathematical bridge: the kinetic energies are related by the equation $E_P = 2E_Q = 2E_R$.

Let's use the first part of this relation, $E_P = 2E_Q$, to solve for our unknown incident energy $E_1$. By substituting our raw equations into this relation, we get:

$$E_1 - 4.0 = 2(E_1 - 4.5)$$

Now, we must carefully expand the right side of the equation. A common pitfall here is forgetting to distribute the factor of 2 to both terms inside the parenthesis.

$$E_1 - 4.0 = 2E_1 - 9.0$$

Next, we rearrange the terms to isolate $E_1$. Subtracting $E_1$ from both sides and adding $9.0$ to both sides yields:

$$9.0 - 4.0 = 2E_1 - E_1$$

$$E_1 = 5.0 	ext{ eV}$$

We have successfully determined that the first light source, which illuminates metals $P$ and $Q$, has an energy of $5.0 	ext{ eV}$.

### Unlocking the Final Metal

With $E_1$ in hand, the rest of the problem unravels like a puzzle. Let's find the kinetic energy of the electrons emitted from metal $Q$.

$$E_Q = 5.0 - 4.5 = 0.5 	ext{ eV}$$

Returning to our given relation, we know that $2E_Q = 2E_R$. Dividing both sides by 2 simply tells us that the kinetic energy of electrons from metal $R$ is identical to those from metal $Q$:

$$E_R = E_Q = 0.5 	ext{ eV}$$

Now, we shift our entire focus to the final metal, $R$. We know its work function is $\phi_R = 5.5 	ext{ eV}$, and we just discovered that its emitted electrons have a maximum kinetic energy of $E_R = 0.5 	ext{ eV}$.

We apply Einstein's Photoelectric Equation one last time for metal $R$:

$$E_R = E_2 - \phi_R$$

$$0.5 = E_2 - 5.5$$

To find the energy of the second light source, $E_2$, we simply add the work function to the kinetic energy:

$$E_2 = 0.5 + 5.5$$

**$$E_2 = 6.0 	ext{ eV}$$**

### Conclusion

The energy of the incident photon used for metal $R$ is exactly $6.0 	ext{ eV}$. This problem is a brilliant exercise in algebraic substitution and physical reasoning. It requires you to keep track of different physical states—different metals and different light sources—and link them together using a fundamental conservation law. Always remember to clearly define your variables and methodically apply the governing equations to each distinct scenario.

Enter Numerical Value:

Visualized Solution

Experimental Setup

Einstein's Photoelectric Equation

Equations for Metals $P$ and $Q$

Applying the Kinetic Energy Relation

Solving for $E_{1}$

Calculating $E_{Q}$ and $E_{R}$

Equation for Metal $R$

Solving for $E_{2}$

Conclusion

The Sigma Insight: Photoelectric Effect

The Master Equation

Setting Up the Algebra

The Crucial Link

Unlocking the Final Metal

Conclusion

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

In a photoelectric effect experiment, the threshold wavelength of light is $380 nm$ . If the wavelength of incident light is $260 nm$ , the maximum kinetic energy of emitted electrons will be Given, $E (in eV) = \frac{1237}{λ (in nm)}$

Wavelengths belonging to Balmer series lying in the range of $450 nm$ to $750 nm$ were used to eject photoelectrons from a metal surface whose work function is $2.0 eV$ . Find (in eV) the maximum kinetic energy of the emitted photoelectrons. (Take $h c = 1242 eV nm$ .)

Enter Numerical Value:

Visualized Solution

Experimental Setup

Einstein's Photoelectric Equation

Equations for Metals .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Q

Applying the Kinetic Energy Relation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E1​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } EQ​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ER​

Equation for Metal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E2​

Conclusion

The Sigma Insight: Photoelectric Effect

The Master Equation

Setting Up the Algebra

The Crucial Link

Unlocking the Final Metal

Conclusion

Similar Questions

Equations for Metals $P$ and $Q$

Solving for $E_{1}$

Calculating $E_{Q}$ and $E_{R}$

Equation for Metal $R$

Solving for $E_{2}$