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Asked by 3047-manvir | 06 Jul, 2022, 07:25: AM
answered-by-expert Expert Answer
Doppler broadening of spectral lines emitted by a high temperature gas is given as
 
begin mathsize 14px style fraction numerator increment lambda over denominator lambda end fraction space equals space 2 over c square root of 2 ln 2 fraction numerator k T over denominator m end fraction end root end style
where Δλ is change in wavelength,  λ is wavelength of spectral line emitted by atom at rest ,
c is speed of light, k is boltzman constant, T is absolute temperature of gas and m is mass of atom
 
begin mathsize 14px style fraction numerator increment lambda over denominator lambda end fraction space equals space fraction numerator 2 over denominator 3 cross times 10 to the power of 8 end fraction square root of fraction numerator ln 4 cross times 1.381 cross times 10 to the power of negative 23 end exponent cross times 1000 over denominator 4 cross times 1.660 space cross times 10 to the power of negative 27 end exponent end fraction end root space equals space 1.132 space cross times space 10 to the power of negative 5 end exponent end style

 
Hence change in wavelength Δλ = 706.52 × 10-9 × 1.132 × 10-5 =  8 × 10-12 m = 8 pm
Answered by Thiyagarajan K | 06 Jul, 2022, 10:52: AM
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plz solve it
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Asked by 3047-manvir | 06 Jul, 2022, 07:26: AM
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plz solve it
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