Two hydrogen like atoms A and B having `(N)/(Z)` ratio equal to 1 have atomic number `Z_(A)` and `Z_(B)` A is moving, while B is stationary and both are in ground state. The minimum kinetic energy which A must have to have an inelastic collision with B such that both A and B getting excited is `221 e V`. If situation is reversed and B is made to stike A which is stationary and both in ground state then minimum kinetic energy which B must have to make an inelastic coliision with both getting excited is `331.5 e V`. Assume mase of proton = mass of neutron.
In which of the following case energy of emitted photon will be same (Neglecting recoil of atom)

Options

A. A makes a transition from n=2 to 1 and B makes a transition from n = 4 to 2
B. A make a transition from n = 4 to 2 and B makes a transition from n = 6 3.
C. A makes a transition from n = 2 to 1 and B makes a trasition from n = 6 to 3
D. A makes a transition from n = 6 to 3 and B makes a transition from n=8 to 4

Answer

B

Explanation

`DeltaE = 13.6 ((1)/((n_(f)/(Z))^(2))-(1)/((n_(1)/(Z))^(2)))`