Question

When light of frequency v is thrown on a metal surface with threshold frequency $v_{0}$, photo-electrons are emitted with maximum kinetic energy = 1.3 × $10^{-18}$J. If the ratio, v : $v_{0}$ = 3 : 1, calculate the threshold frequency $v_{0}$.

Answer 1

Given, Kinetic energy is maximum when emitting the photoelectrons, we have to find the threshold frequency.

                          ${ v }_{ 0 }$= Threshold frequency

                          kinetic energy =$1.3×10^{-18}J$

Also v:${ v }_{ 0 }$ = 1 : 3 which implies v=$\frac { { v }_{ 0 } }{ 3 }$ substitute in K.E formula

                          K.E = $h(v-{ v }_{ 0 })$ where h is planck’s constant.

Substuiting v =$\frac { v_{ o } }{ 3 }$ from the ratio given in question in K.E = $h(v-v_{ o })$

we get $v_{ o } = \frac { -3 \times K.E }{ h2 }$

    ${ v }_{ 0 }=\frac { -3\times 1.3\times { 10 }^{ -18 } }{ 2\times 6.626\times { 10 }^{ -34 } }                          =2.94\times { 10 }^{ 15 } Hz$

Answer 2

HEY MATE....

$\huge{v_{0}=2.94×10_^{15}Hz}.$