The time of revolution of an electron around a nucleus of charge $Ze$ in nth Bohr's orbit is directly proportional to:
A. $n$
B. \[\dfrac{{{n^3}}}{{{Z^2}}}\]
C. $\dfrac{{{n^2}}}{Z}$
D. $\dfrac{Z}{n}$
Answer
612.9k+ views
HintElectrons revolve around the nucleus because the centripetal force is balanced by the electrostatic force of attraction between electron and positively charged nucleus. To find time of revolution of an electron, first need to find the radius and velocity of revolution of electron around the nucleus. Use the Bohr’s postulate of quantization of angular momentum to find the radius and velocity of revolution.
Complete step-by-step solution:
Coulomb force of attraction between the nucleus of charge
$Ze$ and electron revolving in the orbit of radius ${r_n}$ id given by
${F_n} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{Ze \times e}}{{{r_n}^2}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}^2}}$ -(i)
This force provides the necessary centripetal force for the electron to move in a circular orbit of radius ${r_n}$with a speed ${v_n}$.
i.e., $\dfrac{{m{v_n}^2}}{{{r_n}}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}^2}}$ or $m{v_n}^2 = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}}}$ -(ii)
According to Bohr’s postulate of quantization of angular momentum
${L_n} = m{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$
i.e., ${v_n} = \dfrac{{nh}}{{2\pi m{r_n}}}$ -(iii)
Substituting eqn. (iii) in eqn. (ii), we get,
$m{\left( {\dfrac{{nh}}{{2\pi m{r_n}}}} \right)^2} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}}}$ or $\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{r_n}}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}}}$
i.e., ${r_n} = \dfrac{{{n^2}{h^2}{ \in _0}}}{{\pi mZ{e^2}}}$ -(iv)
Substituting eqn. (iv) in eqn. (iii)
${v_n} = \dfrac{{Z{e^2}}}{{2h{ \in _0}n}}$ -(v)
Time period of revolution of electron is given by:
$T = \dfrac{{2\pi {r_n}}}{{{v_n}}}$
Hence, $T\alpha \dfrac{{{r_n}}}{{{v_n}}}$
From eqn. (iv) ${r_n}\alpha \dfrac{{{n^2}}}{Z}$ and from eqn. (v) ${v_n}\alpha \dfrac{Z}{n}$
Hence $T\alpha \dfrac{{{n^3}}}{{{Z^2}}}$.
So, the correct option is B.
Note:- Bohr’s postulates are only applicable on Hydrogen like atom means atoms with only one electron in their valence shell. Bohr’s gives the postulates about hydrogen atoms but these are also applicable hydrogen like atoms.
Complete step-by-step solution:
Coulomb force of attraction between the nucleus of charge
$Ze$ and electron revolving in the orbit of radius ${r_n}$ id given by
${F_n} = \dfrac{1}{{4\pi { \in _0}}}\dfrac{{Ze \times e}}{{{r_n}^2}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}^2}}$ -(i)
This force provides the necessary centripetal force for the electron to move in a circular orbit of radius ${r_n}$with a speed ${v_n}$.
i.e., $\dfrac{{m{v_n}^2}}{{{r_n}}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}^2}}$ or $m{v_n}^2 = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}}}$ -(ii)
According to Bohr’s postulate of quantization of angular momentum
${L_n} = m{v_n}{r_n} = \dfrac{{nh}}{{2\pi }}$
i.e., ${v_n} = \dfrac{{nh}}{{2\pi m{r_n}}}$ -(iii)
Substituting eqn. (iii) in eqn. (ii), we get,
$m{\left( {\dfrac{{nh}}{{2\pi m{r_n}}}} \right)^2} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}{r_n}}}$ or $\dfrac{{{n^2}{h^2}}}{{4{\pi ^2}m{r_n}}} = \dfrac{{Z{e^2}}}{{4\pi { \in _0}}}$
i.e., ${r_n} = \dfrac{{{n^2}{h^2}{ \in _0}}}{{\pi mZ{e^2}}}$ -(iv)
Substituting eqn. (iv) in eqn. (iii)
${v_n} = \dfrac{{Z{e^2}}}{{2h{ \in _0}n}}$ -(v)
Time period of revolution of electron is given by:
$T = \dfrac{{2\pi {r_n}}}{{{v_n}}}$
Hence, $T\alpha \dfrac{{{r_n}}}{{{v_n}}}$
From eqn. (iv) ${r_n}\alpha \dfrac{{{n^2}}}{Z}$ and from eqn. (v) ${v_n}\alpha \dfrac{Z}{n}$
Hence $T\alpha \dfrac{{{n^3}}}{{{Z^2}}}$.
So, the correct option is B.
Note:- Bohr’s postulates are only applicable on Hydrogen like atom means atoms with only one electron in their valence shell. Bohr’s gives the postulates about hydrogen atoms but these are also applicable hydrogen like atoms.
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