Question

Light of 4.5 ev is incident on cesium surface and stopping potential is 0.25 volts so what will be maximum kinetic energy​

Answer 1

Answer:

E=work function + KE max

Explanation:

putting the values will simply give us 0.25 eV

Answer 2

Given:

stopping potential V = 0.25 V

incident light = 4.5 eV

To find:

maximum kinetic energy​ K.E. =?

Step-to-step-explanation:

• The light of 4.5 eV is incident on the cesium surface.
• The maximum kinetic energy of light is the product of incident voltage & stopping potential.

       K.E. = incident voltage ×  stopping potential

              =4.5 ×  0.25

      K.E  = 1.125 eV

• But kinetic energy is always measured in Joule.
• So we have to convert eV to Joule. We know that,

        1 eV = 1.602 × 10⁻¹⁹ joule.

        K.E =  1.125 eV ×  1.602 × 10⁻¹⁹ J

        K.E. = 1.8 × 10⁻¹⁹ J

• Hence, the maximum kinetic energy​ will be 1.8 × 10⁻¹⁹ J.