When photons of energy 4.25 eV strike the surface of a metal A, the ejected photoelectrons have maxi

Q: Why is the de-Broglie wavelength inversely proportional to the square root of kinetic energy?

Because $\lambda = h/p$ and momentum $p = \sqrt{2mK}$, substituting $p$ gives $\lambda = h/\sqrt{2mK}$.

Q: Can the work function be negative?

No, the work function is the minimum energy required to remove an electron from the metal surface, so it must always be a positive value.

Q: What happens if the incident photon energy is less than the work function?

No photoelectric emission will occur, regardless of the intensity of the incident light.

Q: Why do we use the maximum kinetic energy in the photoelectric equation?

The equation $K_{\text{max}} = E - W$ assumes the electron loses no energy to internal collisions before escaping the metal surface.

Q: Is the mass of the electron considered constant in these calculations?

Yes, for these relatively low energies (a few eV), relativistic effects are negligible, so the rest mass of the electron is used.

Question

When photons of energy $4.25 	ext{ eV}$ strike the surface of a metal A, the ejected photoelectrons have maximum kinetic energy $T_A$ expressed in eV and de-Broglie wavelength $\lambda_A$. The maximum kinetic energy of photoelectrons liberated from another metal B by photons of energy $4.70 	ext{ eV}$ is $T_B = (T_A - 1.50 	ext{ eV})$. If the de-Broglie wavelength of these photoelectrons is $\lambda_B = 2\lambda_A$, then

Options:

(A) the work function of A is $2.25 	ext{ eV}$

(B) the work function of B is $4.20 	ext{ eV}$

(C) $T_A = 2.00 	ext{ eV}$

(D) $T_B = 2.75 	ext{ eV}$

Sigma Percentile · Accepted Answer

### Unraveling the Photoelectric Puzzle: A Tale of Two Metals

Imagine two distinct metal surfaces, A and B, being bombarded by photons of different energies. This classic setup is the perfect playground to explore the fascinating interplay between Einstein's photoelectric effect and the de-Broglie wavelength of matter waves.

### The Master Equations

To solve this puzzle, we need two fundamental equations. First, Einstein's photoelectric equation, which tells us that the maximum kinetic energy ($K_{	ext{max}}$) of an ejected electron is the incident photon energy ($E$) minus the metal's work function ($W$):

$$K_{	ext{max}} = E - W$$

Second, we need the de-Broglie wavelength formula, which relates the wavelength ($\lambda$) of a particle to its momentum ($p$) and kinetic energy ($K$):

$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$

### Setting Up the Scene

Let's write down the photoelectric equations for both metals based on the given data. For metal A, the incident energy is $4.25 	ext{ eV}$, and for metal B, it is $4.70 	ext{ eV}$.

$$T_A = 4.25 - W_A$$

$$T_B = 4.70 - W_B$$

We are also given a crucial relationship between the kinetic energies:

$$T_B = T_A - 1.50$$

### The Wavelength Connection

The problem states that the de-Broglie wavelength of electrons from metal B is twice that of electrons from metal A:

$$\lambda_B = 2\lambda_A$$

Substituting the de-Broglie formula into this condition gives us a direct relationship between their kinetic energies:

$$\frac{h}{\sqrt{2m T_B}} = 2 \frac{h}{\sqrt{2m T_A}}$$

Squaring both sides and simplifying, we find:

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

### Solving the Mystery

Now we have a system of two simple equations:
1. $T_A = 4 T_B$
2. $T_B = T_A - 1.50$

Substituting the second equation into the first:

$$T_A = 4(T_A - 1.50)$$

$$T_A = 4T_A - 6.00$$

$$3T_A = 6.00 \implies T_A = 2.00 	ext{ eV}$$

With $T_A$ found, we can easily calculate $T_B$:

$$T_B = 2.00 - 1.50 = 0.50 	ext{ eV}$$

### Finding the Work Functions

Finally, we substitute the kinetic energies back into our initial photoelectric equations to find the work functions of the metals.

For Metal A:

$$W_A = 4.25 - T_A = 4.25 - 2.00 = 2.25 	ext{ eV}$$

For Metal B:

$$W_B = 4.70 - T_B = 4.70 - 0.50 = 4.20 	ext{ eV}$$

**Conclusion:** The work function of A is $2.25 	ext{ eV}$, the work function of B is $4.20 	ext{ eV}$, and $T_A = 2.00 	ext{ eV}$. Therefore, options (a), (b), and (c) are the correct choices.

Select Answer:

Visualized Solution

Visualizing the Setup

Core Principles

Setting up Equations

Using the Wavelength Condition

Relating Kinetic Energies

Solving for Kinetic Energies

Calculating Work Functions

Final Conclusion

The Sigma Insight: Photoelectric Effect

Unraveling the Photoelectric Puzzle

The Master Equations

Setting Up the Scene

The Wavelength Connection

Solving the Mystery

Finding the Work Functions

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

The surface a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is ( $h c = 1240 eV-nm$ )

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 .............. .