Question

What is the energy of the photon emitted by a $B{e}^{3+}$ ion when an electron makes a transition from n = 4 to n = 2?

• A. 13.6 eV
• B. 40.8 eV
• C. 27.2 eV
• D. 54.4 eV

Answer 1

The correct option is B 40.8 eV

We know,
$\Delta E=E_0(\frac{1}{n_1^2}-\frac{1}{n_2^2})Z^2$
where, $E_o$ = Rydberg's energy constant
$n_1$ = lower enrgy shell
$n_2$ = higher energy shell
Z = Atomic number of the element
For $B{e}^{3+}$ , $Z=4$
Putting up the values,
Thus, $\Delta E=13.6\times 16(\frac{1}{2^2}-\frac{1}{4^2})eV$
On solving, 40.8 eV