Question

When a certain metal was irradiated with light of frequency $3.2\times {10}^{16}\text{Hz}$, the photoelectrons emitted had twice the kinetic energy as that of the photoelectrons emitted when the same metal was irradiated with light of frequency $2.0\times {10}^{16}\text{Hz}.$ The value of ${\nu }_{\circ }$ for the metal was found to be $x\times {10}^{16}\text{Hz}$.
Then $x=$

Answer 1

According to photoelectric effect:
$h\nu$ = $w$ + K.E.
where,
h = Planck's constant
h$\nu$ = Energy of photon
$w$ = Work function = $h{\nu }_o$
${\nu }_o$ = threshold frequency
K.E. = Kinetic energy of electrons released from the metal surface

As per the given statements,
h ( 3.2 $\times {10}^{16})=w+2K.E$ ........(1)
h ( 2$\times {10}^{16})=w+K.E.$ ............(2)

Subtracting (2) from (1) we get,
h (1.2 $\times {10}^{16})=K.E.$
$6.626\times {10}^{-34}\times 1.2\times {10}^{16}=7.95\times {10}^{-18}=K.E.$
Put value of K.E in equation (2)
h ( 2$\times {10}^{16})=h{\nu }_o+K.E.$
$(6.626\times {10}^{-34}\times 2\times {10}^{16})=h{\nu }_{\circ }+7.95\times {10}^{-18}$
$\frac{13.25\times {10}^{-18}-7.95\times {10}^{-18}}{6.626\times {10}^{-34}}$ = ${\nu }_{\circ }$
On solving, ${\nu }_o$ = $7.99\times {10}^{15}$ = $0.8\times {10}^{16}\text{Hz}$