Calculate the wavenumber for the longest wavelength transition in the Balmer series of atomic hydrogen.

Solution

straight v to the power of minus space equals space straight R open parentheses 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 parentheses For space Balmer space series comma space space space straight n subscript 1 space equals space 2 comma Hence space straight v to the power of minus space equals space straight R open parentheses 1 over 2 squared minus fraction numerator 1 over denominator straight n subscript 2 superscript 2 end fraction close parentheses Since space straight v to the power of minus space equals space 1 over straight lambda comma space for space straight lambda space to space be space longest space left parenthesis maximum right parenthesis comma straight v to the power of minus space should space be space minimum.

This can only be possible of n₂ is minimum i.e. n₂ = 3. Hence

straight v to the power of minus space equals left parenthesis 1.097 space cross times space 10 to the power of 7 straight m to the power of negative 1 end exponent right parenthesis space open square brackets 1 over 2 squared minus 1 over 3 squared close square brackets equals space 1.097 space cross times space 10 to the power of 7 space cross times space 5 over 36 straight m to the power of negative 1 end exponent

equals space 1.523 space cross times space 10 to the power of 6 straight m to the power of negative 1 end exponent

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