If photons with an energy of 3.2 eV are incident …

Physics Questions

Photons of energy 3.2 eV are incident on a photosensitive surface. It the stopping Potential for the emitted electrons is 1.5 V, the work function for the surface is: A. 1.5 eVB. 1.7 eVC. 3.2eVD. 4.7eV

Short Answer

To find the work function of a photosensitive surface, we use the photon energy of 3.2 eV and the stopping potential of 1.5 V. The maximum kinetic energy of emitted electrons is calculated as 1.5 eV, leading to a work function of 1.7 eV.

Step-by-Step Solution

Step 1: Understand the Problem

To find the work function of a photosensitive surface, we need to consider the energy of the photons and the stopping potential for the emitted electrons. In this case, the energy of the incoming photons is 3.2 eV and the stopping potential is 1.5 V. This information is crucial to determine the maximum kinetic energy of the emitted electrons.

Step 2: Calculate Maximum Kinetic Energy

The maximum kinetic energy (K.E.) of the emitted electrons can be derived using the relationship between the stopping potential and the kinetic energy. It can be expressed with the formula: K.E. = e * V, where e is the charge of an electron. Hence, we calculate:

  • K.E. = 1 * 1.5 eV = 1.5 eV

Step 3: Determine the Work Function

Now, we can find the work function (W) using the formula: W = Photon energy – K.E.. Plugging in the values we have:

  • W = 3.2 eV – 1.5 eV
  • W = 1.7 eV

This results in a work function of 1.7 eV, confirming that the correct option is (b).

Related Concepts

Photon Energy

The energy carried by a photon, which can be quantified in electronvolts (ev)

Stopping Potential

The potential difference needed to stop the flow of emitted electrons, directly related to their kinetic energy

Work Function

A specific energy threshold required to release an electron from a material, calculated as the difference between photon energy and the kinetic energy of emitted electrons.

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