When a surface is irradiated with light of wavelength `4950 Å`, a photocurrent appears which vanishes if a retarding potential greater than 0.6 volt is applied across the phototube. When a second source of light is used, it is found that the critical potential is changed to 1.1 volt.
if the photoelectrons (after emission form the source) are subjected to a magnetic field of 10 tesla, the two retarding potentials would

Answer

`1.9 eV, 4125Å`; no changes

Explanation

`1/2mv_(1)^(2)=eV_(1)=hv_(1)-phi_(0)`
or ` phi_(0)=hv_(1)-eV_(1)=(hc)/(lambda_(1))-eV_(1)`
`:. phi_(0)=(6.63xx10^(-34)xx3xx10^(8))/(4950xx10^(-10))-1.6xx10^(-19)xx0.6`
`=3.04xx10^(-19)J=1.9 eV`
`(hc)/(lambda_(2))=eV_(0)+phi_(0)`
`1.6xx10^(-19)xx1.1+3.04xx10^(-19)`
`=4.8xx10^(-19)J`
`:. lambda_(2)=(hc)/(4.8xx10^(-19))=(6.63xx10^(-34)xx3xx10^(8))/(4.8xx10^(-19))`
`=4125Å`
As the magnetic field does not change the speed of the electrons hence there will be no change in the stopping potential.