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calculate the two longest wavelength of the raditions emitted when hydrogen atoms makes transitions from higher state to n=2 state

pracheta , 6 Years ago
Grade 12th pass
anser 1 Answers
Eshan

Last Activity: 6 Years ago

Dear student,

The two longest wavelengths would be for transitions corresponding to least energy. Hence the two wavelengths correspond to transitions3\rightarrow 2and4\rightarrow 2.

Hence\dfrac{hc}{\lambda_1}=13.6eV(\dfrac{1}{2^2}-\dfrac{1}{3^2})=1.88eV

and\dfrac{hc}{\lambda_2}=13.6eV(\dfrac{1}{2^2}-\dfrac{1}{4^2})=2.55eV

\implies \lambda_1=659.5nm,\implies \lambda_2=486.3nm

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