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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∆ 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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