In a hypothetical atom, a negatively charged particle having a charge of magnitude 3e and mass 3m revolves around a proton. Here, e is the electronic charge and m is the electronic mass. Mass of proton may be assumed to be much larger than that of the negatively charged particle, thus the proton is at rest. This “atom” obeys Bohr’s postulate of quantization of angular momentum, that is $m v r = n (\frac{h}{2 π})$ . It is given that for the first Bohr orbit of hydrogen atom: radius of orbit is $r_{0}$ speed of electron is $v_{0}$ , and total energy is $- E_{0}$ .

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Radius of first orbit of this atom is $\frac{r_{0}}{9}$ .

Radius of first orbit of hypothetical atom is $\frac{r_{0}}{27}$ .

Speed of the revolving particle in the hypothetical atom is $9 v_{0}$ in the first Bohr orbit.

Speed of the revolving particle in the hypothetical atom is $3 v_{0}$ in the first Bohr orbit.

answer is B, C.

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Detailed Solution

If mass of revolving particle is m, and charge q, charge at nucleus Q.
$m v r = n (\frac{h}{2 π})$

$\frac{m v^{2}}{r} = \frac{k (Q) q}{r^{2}}$
Equating $v^{2}$
$\frac{n^{2} {(\frac{h}{2 π})}^{2}}{m^{2} r^{2}} = \frac{k Q q}{m r} \Rightarrow r = \frac{n^{2} {(\frac{h}{2 π})}^{2} n^{2}}{m k Q q}$
Energy of nth orbit
$E_{n} = k + U = \frac{1}{2} m v^{2} - k (\frac{Q q}{r})$
= $\frac{1}{2} k (\frac{Q q}{r}) = \frac{1}{2} \frac{m {(k Q q)}^{2}}{{(\frac{h}{2 π})}^{2} n^{2}}$
For hydrogen atom
Q = e, q = e, m = $m_{e}$ , Q = e, q = e, m = $m_{e}$
For hypothetical atom
Q = e, q = 3e, m = 3 $m_{e}$
Radius of first orbit of this atom

$r_{0}^{1} = \frac{r_{0}}{9}$