represents a graph of most energetic photoelectrons `K_(max)`(in eV) and frequency v for a metal used as cathode in photoelectrons experiment. The threshold frequency of light for the photoelectric emission from the metal is

Options

A. `4xx10^(14)Hz`
B. `3.5xx10^(14)Hz`
C. `2.0xx10^(14)Hz`
D. `2.7xx10^(-19)Hz`

Answer

D

Explanation

From graph, `v=10^(15)Hz`
`K_(max)=3eV=3xx1.6xx10^(-19)J`
As `K_(max)=hv-hv_(0)`
or `v_(0)=v-(K_(max))/(h)=10^(15)-(3xx1.6xx10^(-19))/(6.6xx10^(-34))`
`=(10-7.3)xx10^(14)=2.7xx10^(14)Hz`