Question

in a photoelectric-effect experiment, it is found out that no current flows unless the wavelength is less than 445nm.(a) what is the work function of this material?(b) find the maximum speed of the ejected electrons if the wavelength is changed to (i) 456nm (ii) 342nm

Answer 1

a)

from the given data

Work function of the metal, Wo = h*c/lamda_max

= 6.626*10^-34*3*10^8/(445*10^-9)

= 4.467*10^-19 J or 2.792 eV

b)

i) now Apply,

E_photon = Wo + KEmax

==> KEmax = E_photon - Wo

= h*c/lamda - Wo

= 6.626*10^-34*3*10^8/(456*10^-9) - 4.47*10^-19

= 0 J

so, Vmax = 0

ii)

E_photon = Wo + KEmax

==> KEmax = E_photon - Wo

= h*c/lamda - Wo

= 6.626*10^-34*3*10^8/(342*10^-9) - 4.47*10^-19

= 1.34*10^-19 J

now use, KEmax = (1/2)*m*vmax^2

==> Vmax = sqrt(2*KEmax/m)

= sqrt(2*1.34*10^-19/(9.1*10^-31))

= 5.42*10^5 m/s <<<<----------------Answer