Question

The radius of hydrogen in the ground state is $0.53{{A}^{0}}$. The radius of $L{{i}^{2+}}$ in ground state will be:
a.) \[1.06{{A}^{0}}\]
b.) \[0.265{{A}^{0}}\]
c.) \[0.17{{A}^{0}}\]
d.) \[0.53{{A}^{0}}\]

Answer 1

Hint: The concept to be used is radius in structure of atom. Inside an atom, there are various quantized states in which electrons can perform transition.
- And the radius of atoms of different elements depend on in which state the atom is and the atomic number.
- The radius of hydrogen in ground state is taken as standard to calculate the radius of different states of other elements.

Complete Solution :
The radius of different states of atoms is found in reference to the radius of the ground state of the hydrogen atom. This concept of energy state and finding radius of different elements is due to the Bohr model concepts of quantization of energy inside the structure of an atom.
The radius of an atom with atomic number $Z$ and energy state $n$ is found out to be
$\Rightarrow 0.53\dfrac{{{n}^{2}}}{Z}{{A}^{0}}$

In the given question, we are provided with radius of hydrogen atom in ground state $0.53{{A}^{0}}$ which is taken as a reference
Ground state of $L{{i}^{2+}}$ means $n = 1$
Atomic number of $L{{i}^{2+}}$ = $3$
Putting the values,
Radius of = $0.53\dfrac{{{n}^{2}}}{Z}{{A}^{0}}$

$\Rightarrow 0.53\dfrac{{{\left( 1 \right)}^{2}}}{3}{{A}^{0}}$
$\Rightarrow 0.17{{A}^{0}}$
So, the correct answer is “Option C”.

Note: The transition of electrons in an atom takes place in quantized states. While transitioning, electrons emit the radiation in terms of photons equivalent to the difference between the energy of the two states.
- From the concepts of Bohr model, we can even find the kinetic energy, potential energy, total energy and velocity of an atom in a particular state.