see we can find it with the help of this formula:
delta E=(-RH/nf2)-(-RH/ni2)
where ni and nf stands for initial and final orbit respectively.
also,RH is the rydberg’s constant whose alue is 2.18*10-18J.
SO , on keeping all the values according to the question given....
the solution is as following:
=(2.18*10-18/32)-(-2.18*10-18/12)
=2.18*10-18(1/32-1/12)
=2.18*10-18(1/9-1/1)
=2.18*10-18(1-9/9) {on taking L.C.M of 1 and 9 we get 9}
=2.18*10-18* -8/9
=17.44*10-18/9
=1.94*10-18J
THEREFORE THIS MUCH ENERGY WILL BE EMITTED ..
