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(D) In Bohr's theory,the reduced mass $\mu$ is used to account for the finite mass of the nucleus. The Rydberg constant $R$ is proportional to the red

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$0.05$

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(D) In Bohr's theory,the reduced mass $\mu$ is used to account for the finite mass of the nucleus. The Rydberg constant $R$ is proportional to the reduced mass $\mu = \frac{m_e M}{m_e + M}$,where $m_e$ is the electron mass and $M$ is the nuclear mass.For hydrogen $(H)$,$M_H \approx 2000 m_e$. Thus,$\mu_H = \frac{m_e (2000 m_e)}{m_e + 2000 m_e} = \frac{2000}{2001} m_e$.For deuterium $(D)$,the nucleus has a proton and a neutron,so $M_D \approx 4000 m_e$. Thus,$\mu_D = \frac{m_e (4000 m_e)}{m_e + 4000 m_e} = \frac{4000}{4001} m_e$.Since $\frac{1}{\lambda} \propto \mu$,we have $\lambda \propto \frac{1}{\mu}$.Therefore,$\frac{\lambda_H}{\lambda_D} = \frac{\mu_D}{\mu_H} = \left(\frac{4000}{4001}ight) 	imes \left(\frac{2001}{2000}ight) = 2 	imes \frac{2001}{4001} = \frac{4002}{4001}$.The relative change is $\frac{\Delta \lambda}{\lambda} = \frac{\lambda_H - \lambda_D}{\lambda_H} = 1 - \frac{\lambda_D}{\lambda_H} = 1 - \frac{4001}{4002} = \frac{1}{4002} \approx 0.00025$.Expressed as a percentage,this is $\approx 0.025 \%$. However,using the standard approximation $\frac{\Delta \lambda}{\lambda} \approx \frac{\mu_H - \mu_D}{\mu_D} \approx \frac{1}{2000} = 0.0005$,which leads to $0.05 \%$. Given the options,the intended answer is $0.05 \%$.

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