Question

White light, with frequencies ranging from 4.00 x 10^14 Hz to 7.90 x 10^14 Hz, is incident on a barium surface. Given that the work function of barium is 2.52 eV, find:

(a) the maximum kinetic energy of electrons ejected from this surface.
(b) the range of frequencies for which no electrons are ejected.

Answer 1

0.7515875 eV

`4* 10^(14)\leq f<6.08512* 10^(14)\ Hz`

Step-by-step explanation:

f = Maximum frequency =
`7.9* 10^(14)\ Hz`

h = Planck's constant =
`6.626* 10^(-34)\ m^2kg/s`

W = Work function = 2.52 eV

Converting to Joules

`W=2.52* 1.6* 10^(-19)\\\Rightarrow W=4.032* 10^(-19)\ J`

Maximum photon energy is given by

`E=hf\\\Rightarrow E=6.626* 10^(-34)* 7.9* 10^(14)\\\Rightarrow E=5.23454* 10^(-19)\ J`

Maximum Kinetic energy is given by

`K=E-W\\\Rightarrow K=5.23454* 10^(-19)-4.032* 10^(-19)\\\Rightarrow K=1.20254* 10^(-19)\ J`

Converting to eV

`1.20254* 10^(-19)* (1)/(1.6* 10^(-19))=0.7515875\ eV`

The maximum kinetic energy of electrons ejected from this surface is 0.7515875 eV

`W=hf\\\Rightarrow f=(W)/(h)\\\Rightarrow f=(4.032* 10^(-19))/(6.626* 10^(-34))\\\Rightarrow f=6.08512* 10^(14)\ Hz`

The range of frequencies for which no electrons are ejected is

`4* 10^(14)\leq f<6.08512* 10^(14)\ Hz`