in the photo emissivecell with exicting wavelenght X , THE MAX. kinetic energy of electron is K . if thr exicting wavelength is change to 3/4X the kinetic energy of the fastest emitted electron will be.....

1. equal to 3/4 k

2. 4/3K

3. less than K.

4. more than K.

If the light frequency is greater than a threshold frequency, fo, then a photoelectron will be ejected with kinetic energy up to a maximum value, K, according to the linear formula:

Photoelectric Equation is given by: K = hc/λ – W

(where W is the work function of the metal surface, h and c constants)

If the exciting wavelength is changed by ¾ X then the equation for K_max will be:

K_max1 =   4hc/3X – W

We can see that the K_max1 (that is new max kinetic energy of an electron) is greater than K_max. Hence the kinetic energy of the fastest emitted electron will be more than K.

Answer – 4.