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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). The K.E. of the electrons is

Answer»

`2xx10^(-19)J`
`3xx10^(-19)J`
`1xx10^(-19)J`
`3.18xx10^(-19)J`

Solution :ENERGY of photon `hv=phi_(3)-phi_(2)`
`[-(1.51)+3.4]eV=1.89eV`


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