Question

The radius of the second Bohr orbit for hydrogen atom is:
[ Planck's constant $\left( h \right)=6.6262\times {{10}^{-34}}Js$ ;
Mass of electron $\left( {{m}_{e}} \right)=9.1091\times {{10}^{-31}}kg$ ;
Charge of electron $\left( e \right)=1.60210\times {{10}^{-19}}C$ ;
Permittivity of vacuum $\left( {{\in }_{0}} \right)=8.854185\times {{10}^{-12}}\text{ }k{{g}^{-1}}{{m}^{-3}}{{A}^{2}}$ ]
(A) $1.65\overset{\circ }{\mathop{A }}\,$
(B) $4.76\overset{\circ }{\mathop{A }}\,$
(C) $0.529\overset{\circ }{\mathop{A }}\,$
(D) $2.12\overset{\circ }{\mathop{A }}\,$

Answer 1

Hint : We know that radius of Bohr’s orbit can be calculated by the formula. It is proportional to nth orbit and inversely proportional to square of atomic number. This formula works for single electron species. Also every atom has different energy shells (orbits) known as $K,\text{ }L,\text{ }M,\text{ }N$ and so on. These energy shells are also called levels (first energy level, second energy level and so on).

Complete Step By Step Answer:
Hydrogen atom consists of many imaginary energy shells in which its electron revolves around the nucleus depending on the energy of the electron. The lowest energy state is called ground and when a certain amount of energy is given to the electron, it excites to a higher energy level thus increasing the radius of the atom. Consider an atom that consists of only one electron. Due to the force of attraction and the revolution, the electron possesses an energy (kinetic energy plus potential energy) which depends on the radius of the orbit in which the electron is revolving. Initially, the electron revolves around the nucleus in the first energy level (K shell). This is the lowest energy level. This is also called the ground state of the atom.
$r=\dfrac{{{n}^{2}}{{h}^{2}}}{4{{\pi }^{2}}{{e}^{2}}{{m}_{e}}k}$
Furthermore, this formula is valid for hydrogen-like species i.e. it should have one electron in its shell. The Bohr Theory gives accurate values for the energy levels in hydrogen-like atoms, but it has been improved upon in several respects. Now simply by substituting the values in above equation we get;
$r=\dfrac{{{(2)}^{2}}\times {{\left( 6.6262\times {{10}^{-34}}Js \right)}^{2}}}{4\times {{\left( 3.1416 \right)}^{2}}\times {{\left( 1.60210\times {{10}^{-19}}C \right)}^{2}}\times \left( 9.1091\times {{10}^{-31}}kg \right)\times \left( 9\times {{10}^{9}} \right)}$
On further solving we get;
$r=2.12\times {{10}^{-10}}m$
Finally the radius we get is $r=2.12\overset{\circ }{\mathop{A }}\,$
Therefore correct answer is option D, i.e. the radius of the second Bohr orbit for the hydrogen atom is $2.12\overset{\circ }{\mathop{A }}\,$ .

Note :
Remember that there are certain limitations to Bohr’s theory. It cannot be applied to multi-electron atoms, even one as simple as a two-electron helium atom. Bohr’s model is what we call semi-classical. The orbits are quantized (non-classical) but are assumed to be simple circular paths (classical). As quantum mechanics was developed, it became clear that there are no well-defined orbits; rather, there are clouds of probability.