Modern Physics 1 Question 54
50. A particle of charge equal to that of an electron $-e$, and mass 208 times of the mass of the electron (called a mu-meson) moves in a circular orbit around a nucleus of charge $+3 e$. (Take the mass of the nucleus to be infinite). Assuming that the Bohr model of the atom is applicable to this system, $(1988,6 M)$
(a) derive an expression for the radius of the $n^{\text {th }}$ Bohr orbit.
(b) find the value of $n$ for which the radius of the orbit is approximately the same as that of the first Bohr orbit for the hydrogen atom.
(c) find the wavelength of the radiation emitted when the mu-meson jumps from the third orbit to the first orbit.
(Rydberg’s constant $=1.097 \times 10^{7} m^{-1}$ )
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Answer:
Correct Answer: 50. (a) $113.74 \AA$ (b) 3
Solution:
- If we assume that mass of nucleus $»$ mass of mu-meson, then nucleus will be assumed to be at rest, only mu-meson is revolving round it.
(a) In $n$th orbit the necessary centripetal force to the mu-meson will be provided by the electrostatic force between the nucleus and the mu-meson.
Hence,
$$ \frac{m v^{2}}{r}=\frac{1}{4 \pi \varepsilon _0} \frac{(Z e)(e)}{r^{2}} $$
Further, it is given that Bohr model is applicable to this system also. Hence
Angular momentum in $n^{\text {th }}$ orbit $=\frac{n h}{2 \pi}$
or $\quad m v r=n \frac{h}{2 \pi}$
We have two unknowns $v$ and $r$ (in $n^{\text {th }}$ orbit). After solving these two equations, we get
$$ r=\frac{n^{2} h^{2} \varepsilon _0}{Z \pi m e^{2}} $$
Substituting $Z=3$ and $m=208 m _e$, we get
$$ r _n=\frac{n^{2} h^{2} \varepsilon _0}{624 \pi m _e e^{2}} $$
(b) The radius of the first Bohrs orbit for the hydrogen atom is $\frac{h^{2} \varepsilon _0}{\pi m _e e^{2}}$.
Equating this with the radius calculated in part (a), we get
$$ n^{2} \approx 624 \quad \text { or } \quad n \approx 25 $$
(c) Kinetic energy of atom $=\frac{m v^{2}}{2}=\frac{Z e^{2}}{8 \pi \varepsilon _0 r}$
$$ \begin{array}{ll} \text { and } & \text { potential energy }=-\frac{Z e^{2}}{4 \pi \varepsilon _0 r} \\ \therefore \quad & \text { Total energy } E _n=\frac{-Z e^{2}}{8 \pi \varepsilon _0 r} \end{array} $$
Substituting value of $r$, calculated in part (a),
$$ E _n=\frac{1872}{n^{2}}-\frac{m _e e^{4}}{8 \varepsilon _0^{2} h^{2}} $$
But $-\frac{m _e e^{4}}{8 \varepsilon _0^{2} h^{2}}$ is the ground state energy of hydrogen atom and hence is equal to $-13.6 eV$.
$$ \begin{array}{ll} \therefore & E _n=\frac{-1872}{n^{2}}(13.6) eV=-\frac{25459.2}{n^{2}} eV \\ \therefore \quad E _3-E _1=-25459.2 \frac{1}{9}-\frac{1}{1}=22630.4 eV \end{array} $$
$\therefore$ The corresponding wavelength,
$$ \lambda(\text { in } \AA)=\frac{12375}{22630.4}=0.546 \AA $$
