If a light with frequency `4xx10^(16) Hz` emitted photoelectrons with double the maximum kinetic as are emitted by the light of frequency `2.5xx10^(16) Hz` from the same metal surface, then what is the threshold frequency `(v_(0))` of the metal?

Answer

`1xx10^(16) Hz`

Explanation

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.
`:. 2h v_(o)-h v_(o)=h(5.0xx10^(16)-4xx10^(16))`
`:. h v_(o)=h(1xx10^(16))`
`:.` threshold frequency, `v_(o)=1xx10^(16) Hz`