Question

The radius of the hydrogen atom, when it is in its second excited state, becomes _________ its ground state radius.
A. half
B. double
C. four times
D. nine times

Answer 1

Hint: 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.
Formula used:
${{r}_{n}}=\dfrac{0.53{{n}^{2}}}{Z}\overset{\circ }{\mathop{A}}\,$

Complete step-by-step answer:
Every atom has different energy shells (orbits) known as K, L, M, N and so on. These energy shells are also called levels (first energy level, second energy level and so on).
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.
Suppose an electron is in nth energy level, the radius of this energy orbit is given as ${{r}_{n}}=\dfrac{0.53{{n}^{2}}}{Z}\overset{\circ }{\mathop{A}}\,$
When a particular amount of energy is given to the electron, the energy of the electron increases and it jumps to a higher energy shell. This is called excited state and the radius of orbit increases. When the electron jumps from the first energy shell to second energy shell, it attains the first excited state. If it jumps to third energy it is in a second excited state and so on.
For a hydrogen atom Z=1.
The radius of hydrogen atom when it is in the ground state is ${{r}_{1}}=\dfrac{0.53{{(1)}^{2}}}{1}\overset{\circ }{\mathop{A}}\,=0.53\overset{\circ }{\mathop{A}}\,$
When the hydrogen is in a second excited state, the value of n is equal to 3.
Therefore, the radius of the atom in this state is ${{r}_{3}}=\dfrac{0.53{{(3)}^{2}}}{1}\overset{\circ }{\mathop{A}}\,=0.53\times 9\overset{\circ }{\mathop{A}}\,=9\times {{r}_{1}}$
Therefore, the radius of a hydrogen atom when it is in its second excited state is nine times its radius when it is in its ground state.
Hence, the correct option is D.

Note: Make a note that the formula for radius of an atom given in the solution works only for those atoms or ions which have only one electron in the atom. For example, a hydrogen atom. Another example is ion of helium i.e. $H{{e}^{+}}$.