Question

If a light of frequency $4\times {10}^{16}\text{Hz}$ emits photoelectrons with double the maximum kinetic energy as emitted by a light of frequency $2.5\times {10}^{16}\text{Hz}$ from the same metal surface, what is the threshold frequency $\left({\nu }_0\right)$ of the metal?

• A. $1\times {10}^{16}\text{Hz}$
• B. $1.5\times {10}^{16}\text{Hz}$
• C. $1\times {10}^{15}\text{Hz}$
• D. $1.5\times {10}^{15}\text{Hz}$

Answer 1

The correct option is A $1\times {10}^{16}\text{Hz}$

We konw, $KE=h\nu -h{\nu }_0=h(\nu -{\nu }_0)$
where $\nu$ and ${\nu }_0$ are the incident and the threshold frequencies respectively.
$K{E}_1=h(4\times {10}^{16}\text{Hz})-h{\nu }_0$
and
$K{E}_2=h(2.5\times {10}^{16}\text{Hz})-h{\nu }_0$
As, $K{E}_1=2K.E_2$
$\Rightarrow h(4\times {10}^{16}\text{Hz})-h{\nu }_0=2h(2.5\times {10}^{16}\text{Hz})-2h\left({\nu }_0\right)$
$\Rightarrow (4\times {10}^{16})-(5\times {10}^{16})\text{Hz}=-{\nu }_0$
$\Rightarrow {\nu }_0=1\times {10}^{16}\text{Hz}$