The threshold frequency `v_(0)` for a metal is `6xx10^(14) s^(-1)`. Calculate the kinetic energy of an electron emitted when radiation of frequency `v=1.1xx10^(15) s^(-1)` hits the metal.
Correct Answer - No
Work function `W=hv_(0)=6.626xx10^(-34)xx5.3xx10^(14)=3.5xx10^(-19)J`
As `W gt` Energy of photon, photoelectric effect will not be exhibited.
Correct Answer - `1xx10^(16) Hz`
For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)`
`:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)`
For frequency `4xx10^(16)` Hz,
`h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)`
Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....