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

Answer

B

Explanation

According to Einsteing .s photoelectric equation `K_(max)=h upsilon - 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 = h upsilon - phi_0`
and `e (V_0)/(4) = h (upsilon)/(2) - phi 0`
From (i) and (ii) , we get
`(h upsilon)/(4) - (phi0)/(4) = (h upsilon)/(2) - phi_(0) or phi_0 - (phi0)/(4)= (hupsilon)/(2)-(h upsilon)/(4)`
`3/4 pi_(0) = (h upsilon)/(4) or phi_0 = (hupsilon)/(3)`
`therefore` Threshold frequency, `upsilon_0 = (phi_0)/(h) =(h upsilon)/(3) xx (1)/(h) = (upsilon)/(3)`