Atomic StructureHard

Question

The radius of the hydrogen atom in its ground state is 5.3 × 10–11 m. After collision with an electron it is found to have a radius of 21.2 × 10–11m. The principal quantum number of final state of the atom is

Options

A.2
B.3
C.4
D.16

Solution

$r_{n} = r_{1} \times n^{2} \Rightarrow 21.2 \times 10^{- 11} = 5.3 \times 10^{- 11} \times n^{2} \Rightarrow n = 2$

Create a free account to view solution

View Solution Free
Topic: Atomic Structure·Practice all Atomic Structure questions

More Atomic Structure Questions

The magnitude of spin angular momentum of an electron is given by...Rutherford′s α-particle scattering experiment eventually led to thje conclusion that :...Wavelength of photon which have energy equal to average of energy of photons with λ1 = 4000 Å and λ2 = 6000 Å will be...Sum of number of lone pair of electron and P−O−P linkage present in dimer of P2O5 are :-...Which electronic level would allow the hydrogen atom to absorb a photon but not to emit a photon ?...