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Find out frequency & wave length of a photon emitted during a transition from n=5 to n=2 in H atom. |
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Answer» ∆ E= 2.18x 10-18(1/n1 2 – 1/n2 2 ) = 2.18x 10-18(1/52 – 1/22 ) = -4.58 x 10-19 J Frequency= ∆ E/h = 4.58 x 10-19 J/6.626 x 10-34 Hz =6.91 x 1014 Hz Wave length= C/V0 =3x108 /6.91x1014 =434 nm |
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