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प्रश्न
Find the wavelength of the electron orbiting in the first excited state in hydrogen atom.
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
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, a thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom (~ 10−10 m).
(a) Construct a quantity with the dimensions of length from the fundamental constants e, me, and c. Determine its numerical value.
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(a) will pass through the origin
(b) will be a straight line with slope 4
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(d) will be a circle
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