Chapter 28, Problem 34P

To determine

The wavelength of the photon.

Answer to Problem 34P

  $\text{0}\text{.19}\text{nm}$

Explanation of Solution

Given info:

Number, $\text{n=2}$ to $\text{n=1}$

For iron, $\text{Z}\text{=}\text{26}$

Formula used:

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\text{E}}$

Where, $\text{λ}$ is a wavelength and $\text{E}$ is energy in eV.

Calculation:

With the shielding provided by remaining $\text{n=1}$ electron, use the energies of the hydrogen atom with Z replaced by $\text{Z-1}$

So, the energy of the photon is

  $\begin{array}{l}\text{hf}\text{=ΔE}\\ =-\left(\text{13}\text{.6}\text{eV}\right){\left(\text{26-1}\right)}^{\text{2}}\left[\left(\frac{\text{1}}{{\text{2}}^{\text{2}}}\right)\text{-}\left(\frac{\text{1}}{{\text{1}}^{\text{2}}}\right)\right]\\ \text{=6}{\text{.40×10}}^{\text{3}}\text{eV}\end{array}$

The wavelength of the photon is

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\Delta \text{E}}=\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left(\text{6}{\text{.40×10}}^{\text{3}}\text{eV}\right)}=0.\text{19}\text{nm}$

Conclusion:

The required value is $\text{0}\text{.19}\text{nm}$ .