When a certain metal surface is illuminated wth light of frequency v, the stopping potential for photoelectric current is `V_(0)`. When the same surface is illumiinated by light of frequency `(v)/(2)`, the stopping potential is `(V_(0))/(4)`. The threshold frequency ofr photoelectric emissiohn id

Options

A. `(upsilon )/6 `
B. `(upsilon )/3 `
C. `(2upsilon )/ 3`
D. `(4upsilon )/ 3`

Answer

B

Explanation

According to Einstein.s photoelectric equation
`K_(max) =hupsilon-phi_0`
where `upsilon` is the incident frequency and `phi_0` is the work function of the metal .
As `K_(max) = eV_0` where `V_0` is the stopping potential Therefore,
`eV_0 = hupsilon_(0) = hupsilon - phi_(0) " "...(i)`
and `e(V_0)/4 = h(upsilon)/2 - phi_0 " "...(ii)`
From (i) and (ii) , we get
`upsilon_(0) = phi_0/h=(hupsilon)/3xx1/h=(upsilon)/3`