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प्रश्न
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom.
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उत्तर
For first excited state hydrogen atom is n = 2
Energy of the electron in nth orbit = `−13.6/n^2 eV`
`=13.6/2^2=3.4 eV`
`"Energy" =(hc)/λ=1240/λ`
`3.4 = 1240/λ`
`⇒λ = 364.7 nm`
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संबंधित प्रश्न
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