Question

To remove one electron from helium requires $24.6eV$. and removing its second takes $54.5eV$. The ionization energy of hydrogen is $13.6eV$. When applied to helium -4 by what percentage is equation (11-5) in error due to its ignoring of electronic binding energies?

Short answer

The error in tile binding energy of helium nuclei is $0.000183\%$.

Step-by-step solution

Given data

Energy required removing one electron from helium atom$m=24.6\text{eV}$ .

Energy required remove second electron$=54.5\text{eV}$ .

Formula for Binding energy

For helium- $4$, binding energy is given by, $BE=(Z{m}_H+N{m}_n-M_{\frac{4}{1}x})c^2$.

Calculation for the error in the binding energy of helium nuclei

The value of the dimensions used in binding energy is given as:

$\begin{array}{c}\text{Z}=2\\ m_H=1.007825u\\ N=2\\ m_n=1.008665u\end{array}$

Simplify further as shown below.

$\begin{array}{c}{M}_{\frac{4}{2}x}=4.002603u\\ c=3\times {10}^8\text{m}/\text{s}\end{array}$

Substitute the values in the formula of binding energy.

$\begin{array}{l}BE=[2\left(1.007825u\right)+2\left(1.008665u\right)-4.002603u\mid {(3\times {10}^8\text{m}/\text{s})}^2\\ BE=\left(0.030377u\left(\frac{1.661\times {10}^{-21}kg}{1u}\right)\right){(3\times {10}^8\text{m}/\text{s})}^2\\ BE=4.54\times {10}^{-12}\text{J}\left(\frac{1\text{eV}}{1.6\times {10}^{-19}\text{J}}\right)\\ BE=2.84\times {10}^7\text{eV}\end{array}$

True binding energy for helium nuclei$=24.6\text{eV}+54.5\text{eV}-13.6\text{eV}\times 2=51.9\text{eV}$.

Therefore true binding energy of helium nuclei should be $51.9\text{eV}$ higher.

The error in percentage is given as:

$\begin{array}{l}\text{percentageerror}=\frac{\text{truebindingenergy}}{\text{bindingenergy}}\times 100\%\\ \text{percentageerror}=\frac{51.9\text{eV}}{2.84\times {10}^7\text{eV}}\times 100\%\\ \text{percentageerror}=0.000183\%\end{array}$