Question

Doubly ionized lithium, $\mathit{L}{\mathit{i}}^{\mathbf{2}\mathbf{+}}$absorbs a photon and jumps from the ground state to its n=2level. What was the wavelength of the photon?

Short answer

When, doubly ionized lithium,$L{i}^{2+}$ absorbs a photon and jumps from the ground state to its n=2 level, the wavelength of the photon 13.5 nm.

Step-by-step solution

Finding absorbed energy by the ion:

Whenever an electron absorbs energy enough for it to transition to higher state, it jumps to the higher state, and it loses energy while jumping down to the lower state from the higher state.

As you know that, energy of an electron in its nth orbit is given by

$E_n=-z^2\frac{13.6eV}{n^2}(n=1,2,3...........)$

Where, z

is the atomic number of hydrogen-like atom and n

is the principal quantum number.

Hence, if the ion jumps from ground state to n=2 , the energy will be,

$\begin{array}{rcl}{E}_2-E_1& =& 3^2\frac{-13.6eV}{2^2}{3}^2-\frac{-13.6eV}{1^2}\\ & =& 91.8eV\\ & =& 1.47\times {10}^{-17}J\end{array}$

Finding wavelength of the photon

Energy of the photon is given by,

$E=\frac{hc}{\lambda }.......\left(1\right)$

Here, h is the Planck’s constant, c is the speed of light, and $\lambda$is the wavelength.

The numerical value of hc is,

hc = 1240 eV .nm

Substitute known values into equation (1), and you have

$$\begin{gathered} 91.8eV=\frac{1240eV.nm}{\lambda } \\ \lambda =13.5nm \end{gathered}$$