When two monochromatic lights of frequency, v and (v/2) are incident on a photoelectric metal, their stopping potential becomes (VS/2) and VS respectively. The threshold frequency for this metal is :

Question

When two monochromatic lights of frequency, v and (v/2) are incident on a photoelectric metal, their stopping potential becomes (VS/2) and VS respectively. The threshold frequency for this metal is : (A) (3/2) v (B) 2 v (C) 3 v (D) (2/3) v

Tardigrade Admin · Accepted Answer

hv = W +( v 0/2) e ( h v/2)= W + v 0 e on solving we get, W =3 / 2 h v h o =3 / 2 hv v 0=3 / 2 v

Q. When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V _{S}}{2}$ and $V_{S}$ respectively. The threshold frequency for this metal is :

$\frac{3}{2} v$

$2 v$

$3 v$

$\frac{2}{3} v$

$h v = W + \frac{v _{0}}{2} e$
$\frac{h v}{2} = W + v_{0} e$
on solving we get, $W = 3/2 h v$
$h_{o} = 3/2 h v$
$v_{0} = 3/2 v$

Q. When two monochromatic lights of frequency, v and 2v​ are incident on a photoelectric metal, their stopping potential becomes 2VS​​ and VS​ respectively. The threshold frequency for this metal is :

23​v

2v

3v

32​v