Question

The third line of the Balmer series spectrum of a hydrogen-like ion of atomic number $Z$ equals to $108.5nm$. The binding energy of the electron in the ground state of these ions is $E_B$. Then

• A. $Z=2$
• B. $E_B=54.4eV$
• C. $Z=3$
• D. $E_B=122.4eV$

Answer 1

Answer: (A), (B)

$Z=2$
$E_B=54.4eV$

Third line of Balmer series:
$\lambda =108.5nm=1.085\times {10}^{-7}m$
$\frac{1}{\lambda }=R{Z}^2(\frac{1}{2^2}-\frac{1}{5^2})$
$Z^2=4$
$Z=2$
$E_b=-13.6Z^2/n^2=-54.4eV$ ($n=1$)