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Structure Of Atom

Question
CBSEENCH11005282

Derive the expressions:
left parenthesis straight i right parenthesis space increment straight E space equals space 2.18 space cross times space 10 to the power of negative 18 end exponent straight J space open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus space fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
left parenthesis ii right parenthesis space straight v with bar on top space equals 109677 space open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets cm to the power of negative 1 end exponent

Solution
(i) Suppose the electron is in excited state with n = n2. During emission, the electron drops to a lower energy state with n = n1. The difference between the energies of the initial and final state is given by ∆E.
increment straight E space equals space straight E subscript straight n subscript 2 end subscript minus straight E subscript straight n subscript 1 end subscript
therefore space space increment straight E space equals space fraction numerator negative 2.18 space cross times space 10 to the power of negative 18 end exponent straight J over denominator straight n subscript 2 superscript 2 end fraction minus fraction numerator negative 2.18 space cross times 10 to the power of negative 18 end exponent straight J over denominator straight n subscript 1 superscript 2 end fraction
space space space space space space space space space equals space left parenthesis 2.18 space cross times space 10 to the power of negative 18 end exponent straight J right parenthesis space open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
Since this transition results in the emission of a photon of frequency v with energy hv. We can write
straight E space equals hv space equals space left parenthesis 2.18 space cross times space 10 to the power of negative 18 end exponent straight J right parenthesis space open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
Since space straight h space equals space 6.626 space cross times space 10 to the power of negative 34 end exponent Js
space therefore space space space space space straight v space equals space fraction numerator 2.18 space cross times space 10 to the power of negative 18 end exponent straight J over denominator 6.626 space cross times 10 to the power of negative 34 end exponent Js end fraction open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
space space space space space space space space space equals 3.29 space cross times space 10 to the power of 15 open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets straight s to the power of negative 1 end exponent
In space terms space of space wave space number space left parenthesis straight v with bar on top right parenthesis comma space we space have comma space
left parenthesis straight v with bar on top right parenthesis space equals space straight v over straight c space equals space fraction numerator 3.29 space cross times 10 to the power of 15 straight s to the power of negative 1 end exponent over denominator 3 space cross times space 10 to the power of 10 space cm space straight s to the power of negative 1 end exponent end fraction open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
straight v with bar on top space equals space 109677 space open square brackets fraction numerator 1 over denominator straight n subscript 1 superscript 2 end fraction minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close square brackets
Each spectral line in the emission spectrum corresponds to a particular transition in a hydrogen atom.