When a certain metal surface is illuminated with light of frequency v, the stopping potential for photoelectric current is V0. When the same surface is illuminated by light of frequency (v/2) , the stopping potential is (V0/4). The threshold frequency for photoelectric emission is

Question

When a certain metal surface is illuminated with light of frequency v, the stopping potential for photoelectric current is V0. When the same surface is illuminated by light of frequency (v/2) , the stopping potential is (V0/4). The threshold frequency for photoelectric emission is (A) (v/6) (B) (v/3) (C) (2v/3) (D) (4v/3)

Tardigrade Admin · Accepted Answer

Ist case h v=h v0+e V0 ...(i) (h v/2)=h v0+(e V0/4) ...(ii) Solving Eqs. (i) and (ii), we get V0=(v/3)

$\frac{v}{6}$

$\frac{v}{3}$

$\frac{2 v}{3}$

$\frac{4 v}{3}$

Ist case
$h v = h v_{0} + e V_{0}$ ...(i)
$\frac{h v}{2} = h v_{0} + \frac{e V _{0}}{4}$ ...(ii)
Solving Eqs. (i) and (ii), we get
$V_{0} = \frac{v}{3}$

Q. When a certain metal surface is illuminated with light of frequency v, the stopping potential for photoelectric current is V0.​ When the same surface is illuminated by light of frequency 2v​, the stopping potential is 4V0​​. The threshold frequency for photoelectric emission is

6v​

3v​

32v​

34v​

Ist case hv=hv0​+eV0​ ...(i) 2hv​=hv0​+4eV0​​ ...(ii) Solving Eqs. (i) and (ii), we get V0​=3v​

$\frac{v}{6}$

$\frac{v}{3}$

$\frac{2 v}{3}$

$\frac{4 v}{3}$

Ist case
$h v = h v_{0} + e V_{0}$ ...(i)
$\frac{h v}{2} = h v_{0} + \frac{e V _{0}}{4}$ ...(ii)
Solving Eqs. (i) and (ii), we get
$V_{0} = \frac{v}{3}$