1.

For silver metal, threshold frequency is `1.1 xx 10^(17) s^(-1)`. What is the maximum kinetic energy of the photoelectrons prodcued by shining ultraviolet light of wavelength 1.5 nm on the metal? (Given h = `6.6 xx 10^(-34) J-s, C = 3 xx 10^(8) m//s`]A. `5.94 xx 10^(-17) J`B. `2 xx 10^(-16) J`C. `1 xx 10^(-17) J`D. `8 xx 10^(-15) J`

Answer» Correct Answer - A
`E_("subjected")=hv_(0)+KE`
`thereforeKE = 6.6 xx10^(-34)xx[(3xx10^(8))/(1.5 xx 10^(-9))-1.1xx10^(17)]`
`= 5.94 xx10^(-17)J`


Discussion

No Comment Found

Related InterviewSolutions