Question

An electron in unielectronic species $A^{+x}$ is present in $n^{th}$ orbit. The separartion energy of that electron from $n^{th}$ orbit is $7.65eV$. the frequency of photon required to excite electron from $n^{th}$ orbit to $(n+2{)}^{th}$ orbit is $1.025\times {10}^{15}Hz.$ Mark the correct statement.

• A. Kinetic energy of electron in $(n+2)orbitis1.51eV$
• B. Element $A$ is Beryllium
• C. Value of $xis2$
• D. the value of $(n+1)is4$

Answer 1

The correct option is C Value of $xis2$

Separation energy $=7.65eV$

$-En=7.65$

$\frac{13.6Z^2}{n^2}=7.65...\left(i\right)$

$E_{(n+2)}-En=\frac{hv}{1.6\times {10}^{-19}}$

$\frac{13.6Z^2}{n^2}-\frac{13.6Z^2}{(n+2{)}^2}=\frac{6.63\times {10}^{-34}\times 1.025\times {10}^{15}}{1.6\times {10}^{-19}}$

$7.65-\frac{7.65n^2}{(n+2{)}^2}=4.25$

$\Rightarrow \frac{n^2}{(n+2{)}^2}={\left(\frac{1}{1.5}\right)}^2\Rightarrow n=4$

From equation $\left(i\right)$,

$Z=3$

$A^{+x}=L{i}^{+2}$