Asked by Anonymous
a photon drops from -22 eV to a ground state of -89.2 eV. the frequency the photon emits is?
[1.62]
[1.62]
Answers
Answered by
Elena
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
ε=h•f
f= ε/h= 1.08•10^-17/6.63•10^-34 =1.62•10^16 Hz
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