Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light (v1v2). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1: n then the threshold frequency of the metallic surface is:

Question

Photoelectric emission is observed from a metallic surface for frequencies v1 and v2 of the incident light (v1>v2). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1: n then the threshold frequency of the metallic surface is: (A) (v1-v2/n-1) (B) ( n v1-v2/ n -1) (C) ( n v2-v1/n-1) (D) (v1-v2/n)

Tardigrade Admin · Accepted Answer

Photoelectric equation hv - hv o = K max h(v1-v0)=K1 ldots (1) (1) and h (v2-v0)= K ldots(2) from eq (1) &(2) we get (v1-v0/v2-v0)=(1/n) ⇒ n v1-n v0=v2-v0 nv 1-v2= nv o -v o (( n v1-v2)/( n -1))=v o

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

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

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

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

Q. Photoelectric emission is observed from a metallic surface for frequencies v1​ and v2​ of the incident light (v1​>v2​). If the maximum value of kinetic energy of the photoelectrons emitted in the two cases are in the ration 1:n then the threshold frequency of the metallic surface is:

n−1v1​−v2​​

n−1nv1​−v2​​

n−1nv2​−v1​​

nv1​−v2​​

Photoelectric equation hv−hvo​=Kmax​ h(v1​−v0​)=K1​… (1) (1) and h(v2​−v0​)=K…(2) from eq (1)&(2) we get v2​−v0​v1​−v0​​=n1​ ⇒nv1​−nv0​=v2​−v0​ nv1​−v2​=nvo​−vo​ (n−1)(nv1​−v2​)​=vo​

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

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

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

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