Question

The radius of the first Bohr orbit of a hydrogen atom is $0.53\times {10}^{-10}m$.When an electron collides with this atom which is in its normal state, the radius of the electron orbit in the atom change to $2.12\times {10}^{10}m$.The value of the principal quantum number $n$ of the state to which it is excited is:

• A. $4$
• B. $3$
• C. $2$
• D. $1$

Answer 1

Answer: (C)

$2$

$r_n=a_0{n}^2$
$2.12\times {10}^{-19}=90n^2$
$=0.53n^2$
$n^2=4$
$n=2$
This value of principle quantum number of excited state in $2$