A monochromatic light incident on metal A having threshold frequencynu-not. It emits photo electrons of maximum KE K. Now incident frequency Is made three times and fall on metal B having threshold frequency 2 nu-not. What will maximum KE of photo electrons emitted by metal B?(Answer to this question is given to be-- 3K+h nu-not) Please solve it

I got it and sorry for the last similar question I didn`t figure out this expression as the answer|I`d be using v for incident frequency and v° for threshold one|For metal A;hv=hv°+KK=hv-hv°Multiply throughout by 33K=3hv-3hv°------(1)For metal B;3hv=2hv°+KE(we`re required to find this)KE=3hv-2hv°-----(2)Equation (1)-(2)(Subtraction)3K-KE=3hv-3hv°-3hv+2hv°=-hv°Therefore;KE=3K+hv°