Question

The wavelength of the ${\mathit{K}}_{\mathbf{\alpha }}$line from iron is 193 pm. What is the energy difference between the two states of the iron atom that give rise to this transition?

Short answer

The energy difference between the two states of the iron atom that give rise to this transition is 6.44 keV.

Step-by-step solution

The given data:

The wavelength of the $K_{\alpha }$line from iron, $\lambda =193pm=193\times {10}^{-12}m$

Understanding the concept of magnetic resonance

One electron-volt kinetic energy is acquired by an electron or proton working at a potential change of one volt. In terms of cost and potential difference, the cost formula is

E=eV

Photon energy is the energy carried by a single photon. The amount of energy is directly proportional to the magnetic frequency of the photon and thus, equally, equates to the wavelength of the wave. When the frequency of photons is high, its potential is high.

Formulas:

The kinetic energy gained by the electron is,

$\Delta E=eV$ ….. (1)

Here, e is the charge and V is the accelerating potential difference.

The energy of the photon due to Planck’s relation is,

$E=hf$

$E=\frac{hc}{\lambda }$ ….. (2)

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

Calculation of the energy difference between two states of the iron atom:

Consider the known data as below.

The Plank’s constant,$h=6.63\times {10}^{-34}J\cdot s$

The speed of light,$c=3\times {10}^8\frac{m}{s}$

The charge,$e=1.6\times {10}^{-19}\frac{J}{eV}$

Using the given data in equation (1), the energy difference for the line between the two states of the iron atom as follows:

$$\begin{gathered} ∆E=\frac{\left(6.63\times {10}^{-34}J.s\right)\left(3\times {10}^{8}m/s\right)}{\left(193\times {10}^{-12}m\right)\left(1.6\times {10}^{-19}J/eV\right)} \\ =6.44keV \end{gathered}$$

Hence, the value of the energy difference is 6.44 keV.