What amount of energy should be added to an electron to reduce its de Broglie wavelength `lamda_1=550nm` incident on it, causing the ejection of photo-electrons for which the stopping potential is `V_(s1)=0.19V`. If the radiation of wavelength `lamda_2=190nm` is now incident on the surface, (a) calculate the stopping potential `V_(S2)`, (b) the work function of the surface, and (c) the threshold frequency for the surface.

Explanation

For the first wavelength : `eV_(s1)=hupsilon-W` ...(i)
`W=`work function:
`eV_(s1)=hupsilon_2-W` ...(ii)
Subtracting, we get `V_(S2)-V_(S1)=(h)/(e)(upsilon_2-upsilon_1)` or
`V_(S2)=V_(S1)+(hc)/(e)((1)/(lamda_2)-(1)/(lamda_1))`
`V_(S1)+(hc)/(e)((lamda_1-lamda_2)/(lamda_1lamda_2))`
`=0.19+1240((550-190)/(190xx550))=4.47V`
b. From Eq. (i):
`W=(hc)/(lamda_1)-eV_(S1)`
Work function (in eV)
`(W)/(e)=(hc)/(elamda_1)-V_(S1)=(1240)/(550)-0.19=2.07eV`
c. Threshold frequency:
`V_0=(W)/(h)=(2.07xx1.6xx10^(-19))/(6.62xx10^(-34))=5xx10^(14)Hz`