For the 2P → 1S transition in the hydrogen atom calculate ω. Assuming the
spontaneous emission lifetime of the 2P state to be 1.6 ns, calculate the Einstein B
coefficient. Assume n0 ≈1.
We know that energy of the photon during 2p→1s2p\rightarrow1s2p→1s transition ℏω=Eo4−Eo\hbar\omega =\frac{E_o}{4}-E_oℏω=4Eo−Eo
⇒ℏω=−3Eo4\Rightarrow \hbar\omega =\frac{-3E_o}{4}⇒ℏω=4−3Eo
⇒ℏω=10.2eV\Rightarrow \hbar\omega =10.2eV⇒ℏω=10.2eV
⇒ω=10.2×1.6×10−196.6×10−34J/J.s\Rightarrow \omega=\frac{10.2\times 1.6\times 10^{-19}}{6.6\times 10^{-34}}J/J.s⇒ω=6.6×10−3410.2×1.6×10−19J/J.s
⇒ω=2.47×1015rev/s\Rightarrow \omega =2.47\times 10^{15} rev/s⇒ω=2.47×1015rev/s
B=πe2ϵoℏ=3.14×1.6×1.6×10−388.85×6.6×10−46B=\frac{\pi e^2}{\epsilon_o \hbar}=\frac{3.14\times 1.6\times 1.6\times10^{-38}}{8.85\times 6.6\times 10^{-46}}B=ϵoℏπe2=8.85×6.6×10−463.14×1.6×1.6×10−38 T
=0.137×108T=0.137\times 10^{8}T=0.137×108T
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