Chapter 28, Problem 35P

To determine

The wavelength and reason for not expecting perfect agreement.

Answer to Problem 35P

The wavelength is $\text{0}\text{.061}\text{nm}$

Reason: Partial shielding provided by the $\text{n = 2}$ shell

Explanation of Solution

Given info:

Number, $\text{n = 3}$ to $\text{n=1}$

For molybdenum, $\text{Z}\text{=}\text{42}$

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 the wavelength and $\text{E}$ is energy in eV.

Calculation:

Suppose that the shielding is provided by the 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

  $\text{hf}\text{=ΔE}\text{=}-\left(\text{13}\text{.6}\text{eV}\right){\left(\text{42-1}\right)}^{\text{2}}\left[\left(\frac{\text{1}}{3^{\text{2}}}\right)\text{-}\left(\frac{\text{1}}{{\text{1}}^{\text{2}}}\right)\right]\text{=2}{\text{.03×10}}^{\text{4}}\text{eV}$

The wavelength of the photon is

  $\text{λ=}\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\Delta \text{E}}=\frac{\left(\text{1}\text{.24}{\text{×10}}^{\text{3}}\text{eV}\cdot \text{nm}\right)}{\left(\text{2}{\text{.03×10}}^{\text{4}}\text{eV}\right)}=0.\text{061}\text{nm}$

The reason for not expecting perfect agreement: There is some partial shielding provided by the $\text{n = 2}$ shell, which was ignored while replacing $\text{Z}$ by $\text{Z-1}$ .

Conclusion:

The wavelength is $\text{0}\text{.061}\text{nm}$ .