A 40 W ultraviolet light source of wavelength `2480 A`. Illuminates a magnesium (mg) surface placed 2 m away. Determine the number of photons emitted from the surface per second and the number incident on unit area of Mg surface per second . The photoelectric work function for Mg is `3.68eV`. Calculate the kinetic energy of the fastest electrons ejected from the surface. Determine the maximum wavelength for which the photoelectric effects can be observed with Mg surface.

Explanation

`P =40 W, lambda =2480 Å =248 nm =248xx10^(-9)m`

`P=n(hc)/(lambda)`
`n=(Plambda)/(hc) =(40xx248xx10^(-9))/(6.6xx10^(-34)xx3xx10^(8))`
`= 5xx10^(19)`
where n: number of photons emitted / sec.
Number of photons / sec / `m^(2)` at distance 2 m from source
`n' =(n)/(4pi r^(2)) =(5xx10^(19))/(4pi(2)^(2)) =9.9xx10^(17)`
Energy of incident photon
`E=(1242)/(lambda(nm))eV =(1242)/(248) =5 eV`
`E=phi + K_(max)`
`5 =3.68 + K_(max) implies K_(max) =1.32 eV`
`phi =(hc)/(lambda_(0))`
`lambda_(0) =(1242)/(phi(eV))nm =(1242)/(3.68)`
`=337.5 nm`