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Deutrium was discoveredin 1932byHarold Urey by measuring the small changein wavelength for a particulartransition in .^(1)H and .^(2)H. This isbecause, the wavelength of transition depend to a certainextent on thenuclear mass. If nuclear motion is takeninto account, then the electrons and nucleus revolve around their common centre of mass. Sucha system is equivalent to a single particle with a reduced mass mu,revolvingaround the nucleus at a distance equal tothe electron -nucleusseparation. Here mu = m_(e) M//(m_(e)+M), where M is the nuclear mass and m_(e) is the electronic mass. Estimatethe percentage difference in wavelength for the 1st line of the Lyman seriesin .^(1)H and .^(2)H. (mass of .^(1)H nucleus is 1.6725 xx 10^(-27) kg, mass of .^(2)H nucleus is 3.3374 xx 10^(-27) kg, Mass of electron = 9.109 xx 10^(-31) kg.) |
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Answer» `2.7xx10^(-1)%`. We KNOW that `=1/lambda=R[1/n_f^2-1/n_i^2]` As `n_i` and `n_f`are fixed for by MASS series for hydrogen and deuterium. `lambda prop 1/R " " or " " lambda_D/lambda_H=R_H/R_D`..(i) `R_H=(m_e e^4)/(8 epsilon_0ch^3)=(mu_H e^4)/(8 epsilon_0 ch^3) IMPLIES R_D=(m_e e^4)/(8 epsilon_0ch^3)=(mu_H e^4)/(8 epsilon_0 ch^3)` `therefore R_H/R_D=mu_H/mu_D`...(ii) From equation (i) and (ii) `R_D/R_H=mu_H/mu_D`...(iii) Reduced mass of hydrogen , `mu_H=m_e/(1+m_e//M)=m_e(1-m_e/M)` Reduced mass of deuterium `mu_D=(2M.m_e)/(2M(1+m_e/(2M)))=m_e(1-m_e/(2M))` where M is mass of proton `mu_H/mu_D=(m_e(1-m_e/M))/(m_e(1-m_e/(2M)))=(1-m_e/M)(1-m_e/(2M))^(-1)` `=(1-m_e/M)(1+m_e/(2M))` `mu_H/mu_D=(1-m_e/(2M))` `mu_H/mu_D=(1-1/(2xx1840))=0.99973`...(iv) `(because M=1840 m_e)` From (iii) and (iv) `lambda_D/lambda_H=0.99973 , lambda_D=0.99973 lambda_H` Percentage difference in wavelength of Lyman series in `.^1H` (hydrogen ) and `.^2H` (deuterium) =`((lambda_H-lambda_D)/lambda_H)xx100` `=(1-0.99973 ) x 100 =0.027 %=2.7xx10^(-2)%` |
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