Question

Question #b6ebc

Answer 1

We know that minimum energy required to eject an electron from the surface is called the photoelectric work function `phi_0`.

In case the wave-length `lambda` of the light falling on the photo emitting surface is less than the cut-off wavelength `lambda_0`, rest of the energy of the photon appears as the maximum kinetic energy `E_(k_max)` of ejected photo electron.

As such we have the expression

`hc/lambda=phi_0+E_(k_max)` .......(1)
where `h` is Planck's Constant `=6.626xx10^-34" J"cdot"s"`
or`=4.136xx10^-15" eV"cdot"s"`, `c` is Velocity of light in vacuum`=3xx10^8" m"cdot"s"^-2`
also `phi_0=hc/lambda_0` .......(2)

(a) Using the given and stated values we get work function for tungsten as
`4.136xx10^-15xx(3xx10^8)/(250xx10^-9)=phi_0+0.47`
`=>phi_0=4.96-0.47`
`=>phi_0=4.49" eV"`
(b) Using equation (2) we get
`4.49=(4.136xx10^-15xx(3xx10^8))/lambda_0`
`=>lambda_0=276" nm"`, rounded to units place.

.-.-.-.-.-.-.-.-.-.-.-.

We have made use of Einstein's equation for energy of a photon
`E=hnu`
and wave equation
`c=nulambda`
to write expression for energy of a photon as
`E=hc/lambda`