Photoelectric emission is observed from a surface for frequencies v1, and v2 of the incident radiation (v1, v2). If the maximum kinetic energies of the photoelectrons in two cases are in ratio 1: K then the threshold frequency vo is given by

Question

Photoelectric emission is observed from a surface for frequencies v1, and v2 of the incident radiation (v1, > v2). If the maximum kinetic energies of the photoelectrons in two cases are in ratio 1: K then the threshold frequency vo is given by (A) (v2-v1/K-1) (B) (Kv1-v2/K-1) (C) (Kv2-v1/K-1) (D) (v2-v1/K)

Tardigrade Admin · Accepted Answer

KE1=hv1-hvo KE2 =hv2 -hvo As=(kE1/KE2)=(1/K) (hv1-hvo/hv2-hvo)=(1/K) (Kv1-v2)= vo(K-1) vo= (Kv1 -v2/k-1)

Q. Photoelectric emission is observed from a surface for frequencies $v_{1}$ , and $v_{2}$ of the incident radiation ( $v_{1}, > v_{2}$ ). If the maximum kinetic energies of the photoelectrons in two cases are in ratio $1 : K$ then the threshold frequency $v_{o}$ is given by

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

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

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

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

$K E_{1} = h v_{1} - h v_{o}$
$K E_{2} = h v_{2} - h v_{o}$
As $= \frac{k E _{1}}{K E _{2}} = \frac{1}{K}$
$\frac{h _{v_{1}} - h _{v_{o}}}{h _{v_{2}} - h _{v_{o}}} = \frac{1}{K}$
$(K v_{1} - v_{2}) = v_{o} (K - 1)$
$v_{o} = \frac{K v _{1} - v _{2}}{k - 1}$

Q. Photoelectric emission is observed from a surface for frequencies v1​, and v2​ of the incident radiation (v1​,>v2​). If the maximum kinetic energies of the photoelectrons in two cases are in ratio 1:K then the threshold frequency vo​ is given by

K−1v2​−v1​​

K−1Kv1​−v2​​

K−1Kv2​−v1​​

Kv2​−v1​​