Photoelectric threshold of silver is `lamda=3800A`. Ultraviolet light of `lamda=2600A` is incident of a silver surface. Calculate:
a. the value of work function in joule and in eV.
b. maximum kinetic energy of the emitted photoelectrons.
c. the maximum velocity of the photoelectrons.
(Mass of the electrons`=9.11xx10^(-31)`).

Explanation

a. `lamda_0=3800A`
`W=hf_0=h(c)/(lamda_0)=(6.633xx10^(-34)xx3xx10^(8))/(3800xx10^-10)J`
`=5.23xx10^(-19)eV`
b. Incident wavelength `lamda=2600A`. Therefore,
`f="incident frequency"=(3xx10^(8))/(2600xx10^(-10))Hz`
Then,
`KE_(max)=hf-W_0`
`hf=(6.63xx10^(-34)xx3xx10^(8))/(2600xx10^(-10))=7.65xx10^(-19)J=4.78eV`
`KE_(max)=hf-W_0=4.78eV-3.27eV=1.51eV`
c. `KE_(max)=(1)/(2)mv_(max)^(2)`
`impliesv_(max)=sqrt((2KE_(max))/(m))=sqrt((2xx2.42xx10^(-19))/(9.11xx10^(-31)))`
`=0.7289xx10^(6)ms^(-1)`