Question

Radius of H-atom in its ground state is $5.03\times {10}^{-11}m.$ After collision with an electron it is found to have a radius of $20.12\times {10}^{-11}m.$ The principal quantum number ${}^{\prime }{n}^{\prime }$ of the final state of the atom is:

Answer 1

From Bohr's model of the atom:

$r_n=\frac{n^2{h}^2}{Z{k}_e{e}^2{m}_e}$

where $r_n$ is radius of the nth orbit

n is principal quantum number

h is reduced Plank's constant

Z is atomic number

$k_e$ is coulomb's constant

e is electronic charge

$m_e$ is the mass of electron

$r_1=1\times \frac{h^2}{1\times k_e\times e^2\times m_e}=5.03\times {10}^{-11}\Rightarrow \frac{h^2}{k_e\times e^2\times m_e}=5.03\times {10}^{-11}$

$r_n=\frac{{n^{2}h}^2}{k_e{e}^2{m}_e}=20.12\times {10}^{-11}\to n^2=20.12\times {10}^{-11}\times \frac{k_e{e}^2{m}_e}{h^2}=20.12\times {10}^{-11}\times \frac{1}{5.03\times {10}^{-11}}{n}^2=4\to n=2$