InterviewSolution
Saved Bookmarks
InterviewSolution
| 1. |
Two metallic plates P (collector) and Q (emitter) are separated by a distance of 0.10 meter. These are connected through an ammeter without any cell. A magnetic field B exists parallel to the plates. Light of wavelength between `4000Å and 6000Å` fall on the plate whose work function is 2.39eV. Calculate the minimum value of B for which the current registered with ammeter is zero. Use, `h=6.6xx10^(-34)Js, e=1.6xx10^(-19)C` |
|
Answer» Energy of the incident photon of light of wave length `6000Å` is `E=(hc)/lambda=((6.6xx10^(-34))xx(3xx10^(8)))/(6000xx1.6xx10^(-19))eV=2.01eV` Since this value of energy is less than work function `(=2.39eV)` of the metal, hence no photoelectron is emitted for light of wavelength `6000Å` Since the energy of photon of light of wavelength `4000Å` is greater than 2.39 eV , so photoelectric emission takes place with its helt. Maximum K.E. of the emitted photoelectron is `K_(max)=1/2mv_(max)^(2)=(hc)/lambda-phi_(0)= ((6.6xx10^(-34))xx(3xx10^(8)))/(4000xx10^(-10)xx1.6xx10^(-19))-2.39eV` `=3.1-2.39=0.71 eV=0.71xx1.6xx10^(-19)J` `v_(max)=[(2xx0.71xx1.6xx10^(-19))/(9.1xx10^(-31))]^(1//2)=5xx10^(5)m//s` For zero constantl, `(mv^(2))/r=evB or B=(mv)/(er)=((9.1xx10^(-31))xx5xx10^(5))/((1.6xx10^(-19))xx0.1)=2.86xx10^(-5)T` |
|