Question

An ion Mn+ has a single electron. The highest energy line in its emission spectrum occurs at a frequency of 8.225 × 1016 Hz. Identify the ion. (Enter the symbol of the element in the first box, and its charge in the second.)

Answer 1

Answer :  The element is boron and ion is $B^{+4}$

Explanation :

Using Rydberg's Equation:

$\Delta E_n=-(13.61eV)\times Z^2\left(\frac{1}{n_f^2}-\frac{1}{n_i^2} \right )$

Where,

$\Delta E_n$ = change in energy

Z = atomic number

$n_f$ = Higher energy level = $\infty$

$n_i$= Lower energy level = 1

Putting the values, in above equation, we get

$\Delta E_n=-(13.61eV)\times Z^2\left(\frac{1}{\infty^2}-\frac{1}{1^2} \right )$

$\Delta E_n=(13.61eV)\times Z^2$              ............(1)

As we know that:

$\Delta E=h\nu$

$\Delta E=(6.626\times 10^{-34}J.s)\times (8.225\times 10^{16}s^{-1})$

$\Delta E=5.445\time3s 10^{-17}J=\frac{5.445\time3s 10^{-17}}{1.602\time3s 10^{-19}}=340.2eV$    .......(2)

Now equation 1 and 2, we get:

$(13.61eV)\times Z^2=340.2eV$

Z = 4.99 ≈ 5

The element is boron that has atomic number 5.

As it has only one electron the charge on the B is (+4) that is, $B^{+4}$

Thus, the element is boron and ion is $B^{+4}$