The radius of 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 × 10-11 m . What is the principal quantum number n of the final state of atom 7

Question

The radius of 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 × 10-11 m . What is the principal quantum number n of the final state of atom 7 (A)  n = 4  (B)  n = 2  (C)  n = 16  (D)  n = 3

Tardigrade Admin · Accepted Answer

r propto n2 ie, (rf/ri)=((nf/ni))2 ⇒ (21.2 × 10-11/5.3 × 10-11)=[(n/1)]2 ⇒ n2=4 ⇒ n=2

Q. The radius of hydrogen atom in its ground state is $5.3 \times 1 0^{- 11} m$ . After collision with an electron it is found to have a radius of $21.2 \times 1 0^{- 11} m$ . What is the principal quantum number $n$ of the final state of atom 7

$n = 4$

$n = 2$

$n = 16$

$n = 3$

$r \propto n^{2}$
ie, $\frac{r _{f}}{r _{i}} = (\frac{n _{f}}{n _{i}})^{2}$
$\Rightarrow \frac{21.2 \times 1 0 ^{- 11}}{5.3 \times 1 0 ^{- 11}} = [\frac{n}{1}]^{2}$
$\Rightarrow n^{2} = 4$
$\Rightarrow n = 2$

Q. The radius of hydrogen atom in its ground state is 5.3×10−11m . After collision with an electron it is found to have a radius of 21.2×10−11m . What is the principal quantum number n of the final state of atom 7

n=4

n=2

n=16

n=3