Question

The work functions of Caesium (Cs), Potassium (K) and Sodium (Na) are 2.14eV, 2.30 eV and 2.75eV respectively. If incident electromagnetic radiation has an incident energy of 2.20eV, which of these photosensitive surfaces may emit photoelectrons?

• A. Na only
• B. Cs only
• C. Both Na and K
• D. K only

Answer 1

The Correct Option is B

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.

Answer 2

The correct option is (B):  The minimum energy required for a photoelectron to escape from a metal surface is given by its work function. If the energy of the incident photon is greater than the work function of the metal, then photoelectrons will be emitted.

For the given incident energy of 2.20eV, only Caesium (Cs) can emit photoelectrons because its work function (2.14 eV) is less than the incident energy. The work functions of Potassium (K) and Sodium (Na) are greater than the incident energy, so they cannot emit photoelectrons under these conditions.