Photoelectric emission is observed from a metal surface with incident frequencies v1 v2 where v1 v2. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2: 1, then the threshold frequency v0 of the metal is-

Question

Photoelectric emission is observed from a metal surface with incident frequencies v1 v2 where v1> v2. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2: 1, then the threshold frequency v0 of the metal is- (A) 2 v2-v1 (B) 2 v1-v2 (C) (v1-v2/2) (D) v2-v1

Tardigrade Admin · Accepted Answer

h v1-h v0=k E1 dots (1) hv 2- hv o = kE 2 dots(2) ( h (v1-v0)/ h (v2-v0))=(2/1) v1-v0=2 v2-2 v0 ⇒ v0=2 v2-v1

Q. Photoelectric emission is observed from a metal surface with incident frequencies $v_{1}$ & $v_{2}$ where $v_{1} > v_{2}$ . If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio $2 : 1$ , then the threshold frequency $v_{0}$ of the metal is-

$2 v_{2} - v_{1}$

$2 v_{1} - v_{2}$

$\frac{v _{1} - v _{2}}{2}$

$v_{2} - v_{1}$

$h v_{1} - h v_{0} = k E_{1} \dots (1)$

$h v_{2} - h v_{o} = k E_{2} \dots$ (2)

$\frac{h ( v _{1} - v _{0} )}{h ( v _{2} - v _{0} )} = \frac{2}{1}$

$v_{1} - v_{0} = 2 v_{2} - 2 v_{0}$

$\Rightarrow v_{0} = 2 v_{2} - v_{1}$

Q. Photoelectric emission is observed from a metal surface with incident frequencies v1​ & v2​ where v1​>v2​. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2:1, then the threshold frequency v0​ of the metal is-

2v2​−v1​

2v1​−v2​

2v1​−v2​​

v2​−v1​