Deutrium was discovered in 1932 by Harold Urey by measuring the small change in wavelength for a particular transition in `.^(1)H` and `.^(2)H`. This is because, the wavelength of transition depend to a certain extent on the nuclear mass. If nuclear motion is taken into account, then the electrons and nucleus revolve around their common centre of mass.
Such a system is equivalent to a single particle with a reduced mass `mu`, revolving around the nucleus at a distance equal to the electron -nucleus separation. Here `mu = m_(e) M//(m_(e)+M)`, where M is the nuclear mass and `m_(e)` is the electronic mass. Estimate the percentage difference in wavelength for the `1st` line of the Lyman series in `.^(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`.)

Options

A. `2.7xx10^(-1)%`.
B. `2.7xx10^(-2)%`.
C. `3.5xx10^(-2)%`.
D. `3.5xx10^(-1)%`.

Answer

B

Explanation

Let `mu_H` and `mu_D` are the reduced masses of electron for hydrogen and deuterium respectively.
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)%`