Question

The energy of an electron in an excited $L{i}^{+2}$ ion is $-13.6eV$. Hence angular momentum of the electron is given by (in Js).

• A. $1.05\times {10}^{-34}$
• B. $3.15\times {10}^{-34}$
• C. $2.1\times {10}^{-34}$
• D. $4.2\times {10}^{-34}$

Answer 1

The correct option is B $3.15\times {10}^{-34}$

Energy of $e^{-}$ in $n^{th}$ state of atom with atomic no $z$

$=\frac{-13.6z^2}{n^2}eV$

Now A/c to ques

$\frac{-13.6z^2}{n^2}=-13.6$

$(3{)}^2=n^2$

$n=3$

Angular momentum of $e^{-}$ in $n^{th}$ state $=\frac{nR}{2\pi }$

$=\frac{3\times 6.626\times {10}^{-13}}{2\times 3.14}=3.15\times {10}^{-34}$