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)* Z^2\left((1)/(n_f^2)-(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)* Z^2\left((1)/(\infty^2)-(1)/(1^2) \right )`

`\Delta E_n=(13.61eV)* Z^2` ............(1)

As we know that:

`\Delta E=h\\u`

`\Delta E=(6.626* 10^(-34)J.s)* (8.225* 10^(16)s^(-1))`

`\Delta E=5.445\time3s 10^(-17)J=(5.445\time3s 10^(-17))/(1.602\time3s 10^(-19))=340.2eV` .......(2)

Now equation 1 and 2, we get:

`(13.61eV)* 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)`