Answer
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Hint: Bohr's theory explained the stability of the nucleus of an atom by explaining that electrons move in a fixed orbit and each fixed orbit has fixed energy.
Complete step by step solution:
As the electron shifts to discrete orbits away from the nucleus then the force of attraction between the electron and the positively charged nucleus decreases as per the Coulomb’s law of inverse square. The force of attraction between the nucleus of the atom and the electron offers centripetal force.
As centripetal force decreases, so the centrifugal force should also decrease to maintain net force zero. So, the orbiting speed of an electron decreases as it shifts to discrete orbits away from the nucleus. Hence, the statement given is correct.
As the electron moves away from the nucleus of the atom, the radius of the circular orbit of the electron increases. According to Bohr’s postulate the radius of nth orbit is given as,
\[{r_n} = {a_0}{n^2}\],
Here \[{r_n}\] is the radius of the nth orbit and \[{a_0}\] is the radius of the ground state.
So, the radius of allowed orbits of electrons are proportional to the square of the principal quantum number.
According to Bohr's postulate the frequency of the revolution of electrons is inversely proportional to the cube of the principal quantum number. Hence, the statement given is correct. The energy of the electron in nth orbit is given as,
\[{E_n} = \left( { - 13.6eV} \right)\dfrac{{{Z^2}}}{{{n^2}}}\]
So as the electron shifts away from the nucleus, i.e. higher principle quantum number, the energy decreases. Hence, the statement given is not correct.
Therefore, the correct option is A.
Note: The outward centrifugal force on the orbiting electron in the orbit is due to the speed of the electron which is balanced by the inward force of attraction between the electron and the nucleus.
Complete step by step solution:
As the electron shifts to discrete orbits away from the nucleus then the force of attraction between the electron and the positively charged nucleus decreases as per the Coulomb’s law of inverse square. The force of attraction between the nucleus of the atom and the electron offers centripetal force.
As centripetal force decreases, so the centrifugal force should also decrease to maintain net force zero. So, the orbiting speed of an electron decreases as it shifts to discrete orbits away from the nucleus. Hence, the statement given is correct.
As the electron moves away from the nucleus of the atom, the radius of the circular orbit of the electron increases. According to Bohr’s postulate the radius of nth orbit is given as,
\[{r_n} = {a_0}{n^2}\],
Here \[{r_n}\] is the radius of the nth orbit and \[{a_0}\] is the radius of the ground state.
So, the radius of allowed orbits of electrons are proportional to the square of the principal quantum number.
According to Bohr's postulate the frequency of the revolution of electrons is inversely proportional to the cube of the principal quantum number. Hence, the statement given is correct. The energy of the electron in nth orbit is given as,
\[{E_n} = \left( { - 13.6eV} \right)\dfrac{{{Z^2}}}{{{n^2}}}\]
So as the electron shifts away from the nucleus, i.e. higher principle quantum number, the energy decreases. Hence, the statement given is not correct.
Therefore, the correct option is A.
Note: The outward centrifugal force on the orbiting electron in the orbit is due to the speed of the electron which is balanced by the inward force of attraction between the electron and the nucleus.
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