Question

When a certain metal was irradiated with light of frequency $4\times {10}^{16}{s}^{-1}$. The photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectron emitted when the metal was irradiated with light of frequency$2\times {10}^{16}{s}^{-1}$. If the critical frequency of the metal $\left({\nu }_o\right)=x\times {10}^{16}{s}^{-1}$, then find the value of x.
(given, $h=6.6\times {10}^{-34}Js$)
.

Answer 1

According to Einstein photoelectric equation,
$h\nu =h{\nu }_o+KE$
Where, ${\nu }_o$ is the threshold frequency
According to the question,
$h\times 4\times {10}^{16}=h{\nu }_o+3K.E...\left(1\right)$
$h\times 2\times {10}^{16}=h{\nu }_o+K.E...\left(2\right)$

Multiplying '3' in equation (2), we get
$h\times 6\times {10}^{16}=3h{\nu }_o+3K.E...\left(3\right)$

Subtracting equation (1) from equation (3) we get
$h\times 2\times {10}^{16}=2h{\nu }_o$
$\Rightarrow {\nu }_o={10}^{16}{s}^{-1}$

$\therefore$ x=1