1.

Light of two different frequencies, whose photons have energies 2eV and 5eV respectively, successively illuminates a metal of work function 1.0 eV. The ratio of maximum kinetic energy of the emitted photoelectrons will be

Answer»

`2:5`
`1:5`
`1:4`
`1:2`

SOLUTION :As per RELATION `K_(max)=E-phi_(0)`
`(K_(1))/(K_(2))=(E_(1)-phi_(0))/(E_(2)-phi_(0))=((2-1)eV)/((5-1)eV)1:4`.


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