Question

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 Bohr 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 $\lambda$ (given in terms of Rydberg constant R for the hydrogen atom) equal to

• A. 9/5R
• B. 36/5R
• C. 18/5R
• D. 4/R

Answer 1

Answer: (C)

18/5R

Answer 2

The energy of an electron in the nth orbit of hydrogen atom is $E_n \, \propto \frac{m}{n^2}$ For the hypothetical particle whose mass is double than the electron, energy is $E_n \propto \frac{2m}{n^2}$ The wavelength emitted by the transition of electron is $\frac{1}{\lambda} = R \left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$ $\therefore$ Wavelength emitted by the transition of hypothetical particle is $\frac{1}{\lambda'}$ = 2R$\left[ \frac{1}{n_f^2} - \frac{1}{n_i^2} \right]$ Here, $n_f$ = 2 [First excited state] Therefore, longest wavelength will be emitted if the transition of hypothetical particle takes place from $n_i$ = 3 to $n_f$ = 2. $\therefore$ $\frac{1}{\lambda'} = 2R \left[ \frac{1}{2^2} - \frac{1}{3^2} \right] = 2R \left[ \frac{1}{4} - \frac{1}{9} \right] = \frac{2R \times 5}{4 \times 9} = \frac{5R}{18}$ $\therefore$ $\lambda$' = 18/5R