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

Options

A. `1xx10^(14)Hz`
B. `1.5xx10^(14)Hz`
C. `2.1xx10^(14)Hz`
D. `2.7xx10^(14)Hz`

Answer

d

Explanation

From graphs, `v=10^(15)Hz`,
`K_(max)=3eV=3xx1.6xx10^(-19)J`
As, `K_(max)=hv-hv_(0)` or `hv_(0)=hv-K_(max)`
or `v_(0)=v-(K_(max))/h=10^(15)-(3xx1.6xx10^(-19))/(66xx10^(-34))`
`=(10-7.3)xx10^(14)=2.7xx10^(14)Hz`