When a surface is irradiated with light of wavelength `4950Å`, a photocurrent appears which vanishes if a retarding potential appears which vanishes if a retarding potential greater than 1.2 volt is applied across the phototube. When a different source of light is used, it is found that the critical retarding potential is changed to 2.1 volt. Find the work function of the emitting surface and the wavelength of second source. If the photoelectrons (after emission from the surface) are subjected to a magnetic field of 10 tesla, what changes will be observed in the above two retarding potentials. Use `h=6.6xx10^(-34)Js , e=1.6xx10^(-19)C`.

Explanation

(a) `lambda =495 nm`
`E=(1242)/(lambda(nm)) =(1242)/(495) =2.5 eV, K_(max) =eV_(s) =0.6 eV`
`E= phi + K_(max) implies 2.5 =phi + 0.6`
`phi =1.9 eV`
Let wavelength of second source be `lambda'`,
`K'_(max)=eV'_(s) =1.1 eV`
`E' =phi + K'_(max) =1.9 + 1.1 =3 eV`
`lambda' =(1242)/(E'(eV))nm =(1242)/(3) =414nm`
The magnetic force changes direction of velocity not its magnitude, hence no effect on retarding potentials.
(b) The current registered by ammeter will be zero if no electron reaches to plate P. For this, radius of circular path `Rle10cm`
`R=(mv)/(Bq) =(sqrt(2mK_(max)))/(eB)`
`R prop sqrt(K_(max))`
`lambda =400 nm, E =(1242)/(lambda) =(1242)/(400) -3.1 eV`
`E= phi + K_(max) implies 3.1 =2.39 + K_(max)`
`K_(max) =0.71 eV`
`R=(sqrt(2mK_(max)))/(eB)`
`B =(sqrt(2mK_(max)))/(eR) =(1)/(R) sqrt((2mK_(max))/(e^(2)))`
`=(1)/(0.1) sqrt((2xx9.110^(-31)xx0.71)/(1.6xx10^(-19))) =28.4xx10^(-6) T`
`=28.4 mu T`
For wavelength greater than 400 nm, radius of circular path will be smaller than 10 cm.
(c) `B=mu_(0) ni =4pixx10^(-7)xx20xx100xx2.5 =2xx10^(-3) T`
`R=(sqrt(2mK_(max)))/(Bq) =(sqrt(2mK_(max)))/(eB)`
`K_(max) =(e^(2)B^(2)R^(2))/(2m)`
`=(1.6xx10^(-19)xx(2xx10^(-3))xx(1xx10^(-2))^(2)e)/(2xx9.1xx10^(-31))`
`=0.35 exx10^(2) =35 e` joule
`=(35e)/(e) =35 eV`
`phi =0.5 eV`
`E=phi + K_(max) =0.5 + 35 =35.5 eV`
`lambda=(1242)/(E(eV))nm =(1242)/(35.5) =35 nm`