a photon drops from -22 eV to a ground state of -89.2 eV. the frequency the photon emits is?

Question

[1.62]

Answer 1

ε =89.2-22 = 67.2 eV=67.2•1.6•10^-19= =1.08•10^-17 J.
ε=h•f
f= ε/h= 1.08•10^-17/6.63•10^-34 =1.62•10^16 Hz