Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Boh’r atom model and consider all possible transitions of this hypothetical particle to the first excited level. The longest wavelength photon that will be emitted has wavelength λ (given in terms of the Rydberg constant R for the hydrogen atom) equal to

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36/(5R)

18/(5R)

4/R

9/(5R)

answer is C.

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Detailed Solution

$In{\text{ }}hydrogen{\text{ }}atom\,\,{E_n} = - \frac{{Rhc}}{{{n^2}}}$

Also E_n α m ; where m is the mass of the electron. Here the electron has been replaced by a particle whose mass is double of an electron. Therefore, for this hypothetical atom energy in n^th orbit will be given by ${E_n} = - \frac{{2Rhc}}{{{n^2}}}$

The longest wavelength λ_max (or minimum energy) photon will correspond to the transition of particle from $n = {\text{ }}3{\text{ }} to\;n = {\text{ }}2 \Rightarrow \frac{{hc}}{{{\lambda _{\max }}}} = {E_3} - {E_2} = Rhc\left( {\frac{1}{{{2^2}}} - \frac{1}{{{3^2}}}} \right)$

$This \,\,gives\,\,{\lambda _{\max }} = \frac{{18}}{{5R}}.$

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