Question

8. (The next two questions correspond to the same problem.) You irradiate the surface of a metal (with work function of 4.31 eV) with a KrF laser. The laser wavelength is 248 nm. [3 pts] What is the maximum speed of photoelectrons produced in this experiment? (a) o (b) 123 km/sec (c) 373 km/sec (d) 495 km/sec (e) No photoelectrons are produced. 9. [2 pts) Which action would you take if you would like to reduce the maximum speed of the photoelectrons? (a) Use a laser with larger wavelength. (b) Use a laser with shorter wavelength. (c) Increase the intensity of the laser (d) Decrease the intensity of the laser

Answer 1

8) d) 495 km/s

we know,
KE_max = E_photon - Wo

= h*c/lamda - Wo

= 6.625*10^-34*3*10^8/(248*10^-9) - 4.31*1.6*10^-19 (since 1 eV = 1.6*10^-19 J)

= 1.118*10^-19 J

now use, KE_max = (1/2)*m*v_max^2

==> v_max = sqrt(2*KE_max/m)

= sqrt(2*1.118*10^-19/(9.1*10^-31))

= 4.95*10^5 m/s

= 4.95 km/s

9) a) Use a laser with larger wavelength.

when we increase lamda, E_photon decreases.

so, KE_max also decreases.