Question

The energy required to excite an electron from $n$=2 to $n$=3 energy state is $47.2eV.$ The atomic number of the nucleus, around which the electron is revolving, will be

• A. $5$
• B. $10$
• C. $15$
• D. $20$

Answer 1

The correct option is A $5$

The energy of $n^{th}$ is given by:
$E_n=-\frac{Rch{Z}^2}{n^2}$

For hydrogen atom $Z=1$, then for $n=1$ state
$E_1=-\frac{Rch{1}^2}{1^2}=Rch$
$E_1=-(10.96\times {10}^6\times 3\times {10}^8\times 6.6\times {10}^{-34})=-13.6eV$

Hence, energy required to excite an electron from first orbit is
$E_1=13.6eV$ .............(1)

As per the problem for level $n=2$
$E_2=-\frac{Rch{Z}^2}{2^2}$
$E_2=-\frac{Rch{Z}^2}{4}$ ..................(2)

And for $n=3$
$E_3=-\frac{Rch{Z}^2}{3^2}$
$E_3=-\frac{Rch{Z}^2}{9}$ ....................(3)

Also, the energy required to excite an electron from n=2 to n=3 energy state is
$\Delta E=E_3-E_2$
$47.2=-\frac{Rch{Z}^2}{9}-(-\frac{Rch{Z}^2}{4})$ ............ from (2) and (3)
$47.2=Rch{Z}^2[\frac{1}{4}-\frac{1}{9}]$
$47.2=\left(13.6\right)Z^2\left[\frac{5}{36}\right]$ ............. from (1)
$Z^2=24.98\approx 25$
$Z=5$