Question

Imagine an atom made of a proton and a hypothetical particle of double the mass as that of an electron but the same charge. Apply Bohr theory to consider transitions of the hypothetical particle to the ground state. Then, the longest wavelength (in terms of Rydberg constant for hydrogen atom) is

• A. $\frac{18}{2R}$
• B. $\frac{5}{3R}$
• C. $\frac{10}{3R}$
• D. $\frac{18}{5R}$

Answer 1

The correct option is D $\frac{18}{5R}$

The wavelength of the radiation emitted,

$\frac{1}{\lambda }=R[\frac{1}{(n_f{)}^2}-\frac{1}{(n_i{)}^2}]........\left(1\right)$

The rydberg constant,

$R\propto \text{mass of the electron}\left(e\right)$

The rydberg constant for the new hypothetical particle $R^{{}^{\prime }}$ will be,

$\frac{R^{{}^{\prime }}}{R}=\frac{m^{{}^{\prime }}}{m}=\frac{2m}{m}$

$\Rightarrow R^{{}^{\prime }}=2R$

From, (1) we get, For the maximum wavelength,

$n_f=2;n_i=3$

$\frac{1}{{\lambda }_{max}}=2R[\frac{1}{(2{)}^2}-\frac{1}{(3{)}^2}]$

$\frac{1}{{\lambda }_{max}}=2R\left[\frac{5}{36}\right]$

$\Rightarrow {\lambda }_{max}=\frac{18}{5R}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.