Dual Nature Of Radiation And Matter

To solve this problem, we need to relate the given electromagnetic wave of wavelength with the de-Broglie wavelength of the emitted photoelectron. The key points to consider are:

1. The energy of the incident electromagnetic wave can be expressed using the formula for photon energy: , where is Planck's constant, and is the speed of light.
2. Since the work function is negligible, all the energy of the photon is converted into the kinetic energy of the emitted photoelectron.
3. The kinetic energy of the photoelectron can be related to its de-Broglie wavelength using the formula: , where is the mass of the photoelectron and is its velocity.

Now let's deduce the relation step-by-step:

1. From the energy conservation perspective, the kinetic energy ( ) of the emitted electron is given as: .
2. Using the de-Broglie relationship, we have: .
3. By rearranging the above de-Broglie equation for velocity , we get: .
4. Substitute the velocity expression back into the kinetic energy expression: .
5. Simplifying the equation, we obtain: .
6. Rearranging terms to solve for the incident wavelength , we have: .

Thus, the correct relation between the original wavelength and the de-Broglie wavelength is: . This matches the given correct answer.

Step 1: Understanding the Photoelectric Effect:

The photoelectric effect refers to the emission of electrons (photoelectrons) from a photosensitive surface when it absorbs energy from incident light (electromagnetic radiation). The key requirement for this effect to occur is that the energy of the incident photons must be greater than or equal to the work function (the minimum energy needed to release an electron) of the material.

Step 2: Given Information:

• Work functions:
  • Cesium (Cs): 2.14 eV
  • Potassium (K): 2.30 eV
  • Sodium (Na): 2.75 eV
• Incident photon energy: 2.20 eV

Step 3: Analyzing the Options:

Option 1: Na only

Sodium has a work function of 2.75 eV, which is greater than the incident photon energy (2.20 eV). Since the energy of the incident light is less than the work function of sodium, no photoelectron will be emitted from sodium.

Option 2: Cs only

Cesium has a work function of 2.14 eV, which is less than the incident photon energy (2.20 eV). As the incident light has enough energy to overcome the work function, photoelectrons will be emitted from cesium.

Option 3: Both Na and K

Potassium has a work function of 2.30 eV, which is slightly higher than the incident energy (2.20 eV), so no photoelectron will be emitted from potassium. Sodium, as we analyzed earlier, will also not emit photoelectrons. Hence, this option is incorrect.

Option 4: K only

Potassium has a work function of 2.30 eV, which is higher than the energy of the incident light (2.20 eV). Hence, potassium will not emit photoelectrons either. This option is incorrect.

Step 4: Conclusion:

The only material that will emit photoelectrons is Cesium, as its work function is lower than the energy of the incident light.

Answer: The correct answer is Option 2: Cs only.