Question

Photoelectric emission is observed from a metal surface with incident frequencies${\nu }_{1}and{\nu }_2$ where ${\nu }_1>{\nu }_2$. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio $2:1$, then the threshold frequency ${\nu }_o$ of the metal is

• A. ${\nu }_1-{\nu }_2$
• B. $\frac{{\nu }_1-{\nu }_2}{h}$
• C. $2{\nu }_1-{\nu }_2$
• D. $2{\nu }_2-{\nu }_1$

Answer 1

The correct option is D $2{\nu }_2-{\nu }_1$

Applying Einstein's Photoelectric Equation,
$K.E=E-W_o$
$h{\nu }_1=h{\nu }_o+K.E$ ....(i)
$h{\nu }_2=h{\nu }_o+\frac{K.E}{2}$
$2h{\nu }_2=2h{\nu }_o+K.E$ ....(ii)
Subtracting equations (i) from equation (ii) , We get
${\nu }_o=2{\nu }_2-{\nu }_1$