1.

Direction : The questions 57, 58 and 59 are based on following paragraph : Radiation emitted when an electron jumps from n = 3 to n=1 orbit in H_(2) atom, fall on a metal to produce photoelectrons. Electrons fall on the metal surface with maximum K.E. in a direction perpendicular to a magnetic field of 1/320 T in a radius of 10m^(-3). Work function of metal

Answer»

1 eV
2 eV
1.03 eV
3 eV

Solution :`(mv^(2))/(R)=Bev rArr v=(Ber)/(m)`
`K.E.=1//2`
`mv^(2)=1//2 m((Ber)/(m))^(2)=1.378xx10^(-19)Js`
Now `(K.E.)_("max")= hv-phi_(0)`
`:. phi_(0)=hv-(K.E.)_("max")=1.03eV`


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