Question

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-

• A. $2 v_{2}-v_{1}$
• B. $2 v_{1}-v_{2}$
• C. $\frac{v_{1}-v_{2}}{2}$
• D. $v_{2}-v_{1}$

Answer 1

Answer: (A)

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

$hv _{2}- hv _{ o }= kE _{2}\, \dots$(2)

$\frac{ h \left(v_{1}-v_{0}\right)}{ h \left(v_{2}-v_{0}\right)}=\frac{2}{1}$

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

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