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. `1xx10^(14)` Hz
B. `1.5xx10^(15)`Hz
C. `2.1xx10^(14)`Hz
D. `2.8xx10^(14)`Hz

Answer

D

Explanation

According to graph,
for `upsilon=10^(15)Hz,`
`K_("max")=3"eV"=3xx1.6xx10^(-19)"J"`
According to Einstein's photoelectric equation
`K_("max")=hupsilon-hupsilon_(0)" or "hupsilon_(0)=hupsilon-K_("max")`
or `" "upsilon_(0)=upsilon-(K_("max"))/(h)`
`=10^(15)-(3xx1.6xx10^(-19))/(6.63xx10^(-34))`
`=(10-7.2)xx10^(14)=2.8xx10^(14)Hz`