Photoelectric emission is observed from a met | vedclass

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Question

(B) According to Einstein's photoelectric equation,the maximum kinetic energy ${K_{\max }}$ is given by ${K_{\max }} = h
u - h{
u _0}$,where ${
u _

Vedclass · Accepted Answer

$\frac{{k
u _1 - 
u _2}}{{k - 1}}$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy ${K_{\max }}$ is given by ${K_{\max }} = h
u - h{
u _0}$,where ${
u _0}$ is the threshold frequency.For frequency ${
u _1}$,${K_1} = h({
u _1} - {
u _0})$.For frequency ${
u _2}$,${K_2} = h({
u _2} - {
u _0})$.Given the ratio of maximum kinetic energies is ${K_1}:{K_2} = 1:k$,we have $\frac{{{K_1}}}{{{K_2}}} = \frac{1}{k}$.Substituting the expressions,we get $\frac{{h({
u _1} - {
u _0})}}{{h({
u _2} - {
u _0})}} = \frac{1}{k}$.$k({
u _1} - {
u _0}) = {
u _2} - {
u _0}$.$k{
u _1} - k{
u _0} = {
u _2} - {
u _0}$.$k{
u _1} - {
u _2} = k{
u _0} - {
u _0} = {
u _0}(k - 1)$.Therefore,the threshold frequency is ${
u _0} = \frac{{k{
u _1} - {
u _2}}}{{k - 1}}$.

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