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The radiation corresponding to 3 to 2 transition of hydrogen atom falls on a metal surface to produce photoelectrons.These electrons are made to enter a magnetic field of 3xx10^(-4)T.If the radius of the largest circular path followed by these electrons is 10.0 mm the work function of the metal is close to: |
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Answer» 1.1 EV The energy in n=2 orbit `E_(2)=-(13.6)/(4)=-3.4eV` The energy in n=1 orbit `E_(1)=-13.6eV` The energy of EMITTED PHOTONS during the transition `3to2` of electron, `hf=E_(3)-E_(2)` `=-1.51-(-3.4)=3.4-1.51` A moving electron moves in a circle in a magnetic FIELD, `:.(mv^(2))/(r)=Bqv` `:.mv=Bqr` `:.p=Bqr` and `K_(max)=(p^(2))/(2m)=((Bqr)^(2))/(2m)` in joule `:.K_(max)=((Bqr)^(2))/(2mxx1.6xx10^(-19))eV` `=((3xx10^(-4)xx1.6xx10^(-19)xx10^(-2))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19))` `=(2304)/(29.12)xx10^(-2)` `=0.791eV"".......(2)` Now, from photoelectric equation of Einstein, `K_(max)=hf-phi` `:.phi=hf-K_(max)` `=1.89-0.791""` [From equ. (1) and (2) ] `=1.099 eV` `:.phi~~1.1eV` |
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