Question

Photoelectric emission is observed from a metal surface when the metal is subjected to radiations of 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 $v_0$ 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 $2{\nu }_2-{\nu }_1$

$h{\nu }_1=h{\nu }_0+E...\left(i\right)$
where E is the kinetic energy of the electron.
$h{\nu }_2=h{\nu }_0+\frac{1}{2}E$ or $2h{\nu }_2=2h{\nu }_0+E...\left(ii\right)$
Eqn. (i) - Eqn. (ii) gives:
$h{\nu }_1-2h{\nu }_2=h{\nu }_0-2h{\nu }_0$
(or)
${\nu }_1-2{\nu }_2=-{\nu }_0$
or
${\nu }_0=2{\nu }_2-{\nu }_1$