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(C) In a hydrogen atom,the energy of the $n^{th}$ level is given by $E_n = -\frac{Rhc}{n^2}$.Since the energy $E_n$ is directly proportional to the ma

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$18/(5R)$

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(C) In a hydrogen atom,the energy of the $n^{th}$ level is given by $E_n = -\frac{Rhc}{n^2}$.Since the energy $E_n$ is directly proportional to the mass of the orbiting particle $(E_n \propto m)$,and the new particle has a mass $m' = 2m_e$,the energy levels for this hypothetical atom become $E'_n = -\frac{2Rhc}{n^2}$.The longest wavelength photon corresponds to the minimum energy transition,which occurs between $n = 3$ and $n = 2$.The energy difference is $\Delta E = E'_3 - E'_2 = -\frac{2Rhc}{3^2} - (-\frac{2Rhc}{2^2}) = 2Rhc \left( \frac{1}{4} - \frac{1}{9} \right)$.Simplifying this,$\Delta E = 2Rhc \left( \frac{5}{36} \right) = \frac{5Rhc}{18}$.Since $\Delta E = \frac{hc}{\lambda}$,we have $\frac{hc}{\lambda} = \frac{5Rhc}{18}$.Therefore,$\lambda = \frac{18}{5R}$.

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