Question

9. The work function for silver is 4.73 eV. Ultraviolet light of wavelength 116 nm is incident on the cl a silver sample. (a) What is the photon energy of the ultraviolet light in units of eV? (b) What is the maximum kinetic energy of the ejected photoelectrons? (c) What stopping voltage would be required to stop the current of photoelectrons? (5 points) (10 points) (5 points)

Answer 1

ve thalt a) E lev) 124 ev nm b) n oto 5.459 ev 9) stopping potential

Answer 2

SOLUTION :

a.

Photon , a quantum of EM radiation has energy as under :

E = h f = h c / ? 

where,

E = energy of a single photon

h = planck’s constant = 4.14 * 10^(-15) eV.s

f = frequency of the wave

? = wavelength of ultraviolet light.

c = speed of light = 3.0*10^8 m/s 

So,

E = 4.14*10^(-15) * 3.0*10^8 / (116*10^(-9)  eV 

= 10.707 eV (ANSWER).

b.

Maximum K. E. of photoelectron 

= Photon energy - W of silver.

= 10.707 - 4.73

= 5.977 eV (ANSWER).

c.

Stopping voltage too stop the current of potoelecron

= Maximum K.E. of photoelectron / q 

= 5.977 / (- 1.602 * 10^(-19) )  eV/C

= 5.977 / (- 1.602 * 10^(-19)). *( 1.602  * 10^(-19))  volts 

= - 5.977 volts 

= 5.977 volts (magnitude) (ANSWER).