Question

The L absorption edge of a neodymium atom (Z 60) is 1.9 A. How much work is required to ionize a K electron from a neodymium atom? hullan in al In the rest energy of an electro

(write what method and basic logic are used,plz)

Answer 1

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As Ionization energy is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation. It is quantitatively expressed in symbols as:

X + energy → X⁺ + e⁻

where X is any atom or molecule capable of being ionized, X⁺ is that atom or molecule with an electron removed, and e⁻ is the removed electron. This is anendothermic process.

Atomic ionization energy can be predicted by an analysis using electrostatic potential and the Bohr model of the atom, as follows (note that the derivation usesGaussian units).

Consider an electron of charge -e and an atomic nucleus with charge +Ze, where Z is the number of protons in the nucleus. According to the Bohr model, if the electron were to approach and bond with the atom, it would come to rest at a certain radius a. The electrostatic potential V at distance a from the ionic nucleus, referenced to a point infinitely far away, is:

$V={\frac {Ze}{a}}\,\!$

Since the electron is negatively charged, it is drawn inwards by this positive electrostatic potential. The energy required for the electron to "climb out" and leave the atom is:

$E=eV={\frac {Ze^{2}}{a}}\,\!$

This analysis is incomplete, as it leaves the distance a as an unknown variable. It can be made more rigorous by assigning to each electron of every chemical element a characteristic distance, chosen so that this relation agrees with experimental data.

It is possible to expand this model considerably by taking a semi-classical approach, in which momentum is quantized. This approach works very well for the hydrogen atom, which only has one electron. The magnitude of the angular momentum for a circular orbit is:

$L=|\mathbf {r} \times \mathbf {p} |=rmv=n\hbar$

The total energy of the atom is the sum of the kinetic and potential energies, that is:

$E=T+U={\frac {p^{2}}{2m_{e}}}-{\frac {Ze^{2}}{r}}={\frac {m_{e}v^{2}}{2}}-{\frac {Ze^{2}}{r}}$

Velocity can be eliminated from the kinetic energy term by setting the Coulomb attraction equal to the centripetal force, giving:

$T={\frac {Ze^{2}}{2r}}$

Solving the angular momentum for v and substituting this into the expression for kinetic energy, we have:

${\frac {n^{2}\hbar ^{2}}{rm_{e}}}=Ze^{2}$

This establishes the dependence of the radius on n. That is:

$r(n)={\frac {n^{2}\hbar ^{2}}{Zm_{e}e^{2}}}$

Now the energy can be found in terms of Z, e, and r. Using the new value for the kinetic energy in the total energy equation above, it is found that:

$E=-{\frac {Ze^{2}}{2r}}$

At its smallest value, n is equal to 1 and r is the Bohr radius a₀ which equals to ${\frac {\hbar ^{2}}{me^{2}}}$ . Now, the equation for the energy can be established in terms of the Bohr radius. Doing so gives the result:

$E=-{\frac {1}{n^{2}}}{\frac {Z^{2}e^{2}}{2a_{0}}}=-{\frac {Z^{2}13.6eV}{n^{2}}}$

we know the value of Z i.e 60 and absorbtion edge (L) of neodymium atom is 1.9 , put the value in formula and you will get the answer .

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