When photon of energy 4.0 eV strikes the surface of a metal A, the ejected photoelectrons have maxim

Q: Why do we use the formula $p = \sqrt{2mT}$ instead of just $p = mv$?

While $p = mv$ is correct, the problem gives us information about kinetic energy ($T$), not velocity. The formula $T = \frac{p^2}{2m}$ directly relates kinetic energy to momentum, allowing us to substitute $p = \sqrt{2mT}$ into the de-Broglie equation $\lambda = \frac{h}{p}$.

Q: Does the mass of the electron change between metal A and metal B?

No, the mass of an electron ($m$) is a fundamental constant. Because it is the same for electrons ejected from both metals, the $m$ terms cancel out when we equate the two wavelength expressions.

Q: Why did we square the wavelength equation $\frac{h}{\sqrt{2mT_B}} = 2 \frac{h}{\sqrt{2mT_A}}$?

Squaring both sides is a standard algebraic technique to eliminate square roots. It makes solving for the variables $T_A$ and $T_B$ much easier and prevents potential sign errors during manipulation.

Q: Can the work function of a metal ever be negative?

No, the work function represents the minimum amount of energy required to remove an electron from the surface of a metal. Since it is an energy barrier that must be overcome, it is always a positive value.

Q: What would happen if the incident photon energy $E_B$ was less than the work function $\phi_B$?

If the incident photon energy is less than the work function ($E_B < \phi_B$), the photon does not have enough energy to overcome the binding energy of the electron. In this case, no photoelectric emission would occur, and no electrons would be ejected.

Question

When photon of energy $4.0	ext{ eV}$ strikes the surface of a metal $A$, the ejected photoelectrons have maximum kinetic energy $T_A	ext{ eV}$ and de-Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal $B$ by photon of energy $4.50	ext{ eV}$ is $T_B = (T_A - 1.5)	ext{ eV}$. If the de-Broglie wavelength of these photoelectrons $\lambda_B = 2\lambda_A$, then the work function of metal $B$ is

Options:

(A) $4	ext{ eV}$

(B) $2	ext{ eV}$

(C) $1.5	ext{ eV}$

(D) $3	ext{ eV}$

Sigma Percentile · Accepted Answer

Imagine you are conducting a fascinating experiment in a quantum physics lab. You have two different metal surfaces, let's call them Metal $A$ and Metal $B$. You are firing photons at both of them and observing the electrons that get kicked out—a classic demonstration of the photoelectric effect.

For Metal $A$, the incoming photon has an energy of $E_A = 4.0	ext{ eV}$. This photon ejects an electron with a maximum kinetic energy $T_A$ and a de-Broglie wavelength $\lambda_A$. For Metal $B$, you use a slightly more energetic photon with $E_B = 4.5	ext{ eV}$. The ejected electron from this metal has a kinetic energy $T_B$ and a wavelength $\lambda_B$.

### The de-Broglie Connection

To solve this puzzle, we first need to connect the kinetic energy of these electrons to their de-Broglie wavelengths. We know the de-Broglie wavelength $\lambda$ is given by Planck's constant $h$ divided by momentum $p$. Since momentum can be expressed in terms of kinetic energy $T$ as $p = \sqrt{2mT}$, our master formula becomes:

$$\lambda = \frac{h}{\sqrt{2mT}}$$

The problem gives us a very specific relationship between the wavelengths of the electrons from the two metals: $\lambda_B = 2\lambda_A$. Let's substitute our master formula into this condition:

$$\frac{h}{\sqrt{2mT_B}} = 2 \left( \frac{h}{\sqrt{2mT_A}} ight)$$

Notice how the $h$ and the $\sqrt{2m}$ terms cancel out beautifully on both sides. To get rid of the square roots, we square both sides of the equation:

$$\frac{1}{T_B} = \frac{4}{T_A}$$

Cross-multiplying gives us a neat and simple relation: **$T_A = 4T_B$**.

### The Energy Link

We have another crucial piece of information. The problem states that the kinetic energy of the electron from Metal $B$ is $1.5	ext{ eV}$ less than that from Metal $A$. Mathematically, this is written as:

$$T_B = T_A - 1.5$$

Now, let's bring back the relation we just found ($T_A = 4T_B$) and substitute it into this equation:

$$T_B = 4T_B - 1.5$$

Rearranging this to solve for $T_B$, we move the $T_B$ terms to one side:

$$3T_B = 1.5$$

Dividing by $3$, we find that the kinetic energy of the electron from Metal $B$ is exactly **$T_B = 0.5	ext{ eV}$**.

### The Final Photoelectric Equation

We are almost at the finish line. We need to find the work function of Metal $B$, denoted as $\phi_B$. What connects incident photon energy, work function, and kinetic energy? Einstein's famous photoelectric equation!

For Metal $B$, the equation is:

$$E_B = \phi_B + T_B$$

Let's plug in the numbers we know. The incident photon energy $E_B$ is $4.5	ext{ eV}$, and the kinetic energy $T_B$ we just calculated is $0.5	ext{ eV}$:

$$4.5 = \phi_B + 0.5$$

Subtracting $0.5$ from $4.5$, we get the work function:

**$\phi_B = 4.0	ext{ eV}$**

And there we have it! The work function of Metal $B$ is $4.0	ext{ eV}$, which perfectly matches option (a). This is a beautiful problem that elegantly combines the principles of the photoelectric effect with the concept of de-Broglie wavelengths.

Select Answer:

Visualized Solution

\text{Photoelectric Setup}

\text{de-Broglie Wavelength}

\text{Wavelength Condition}

\text{Kinetic Energy Relation}

\text{Energy Difference Condition}

\text{Solving for } T_B

\text{Einstein's Photoelectric Equation}

\text{Calculating Work Function } \phi_B

\text{Conclusion}

The Sigma Insight: Photoelectric Effect

The de-Broglie Connection

The Energy Link

The Final Photoelectric Equation

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