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Expert-verified Found in: Page 518 ### Modern Physics

Book edition 2nd Edition
Author(s) Randy Harris
Pages 633 pages
ISBN 9780805303087 # To remove one electron from helium requires ${24}{.}{6}{e}{V}$. and removing its second takes ${54}{.}{5}{e}{V}$. The ionization energy of hydrogen is ${13}{.}{6}{e}{V}$ . When applied to helium -4 by what percentage is equation (11-5) in error due to its ignoring of electronic binding energies?

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

See the step by step solution

## Step 1: Given data

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

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

## Step 2: Formula for Binding energy

For helium- ${4}$, binding energy is given by, ${B}{E}{=}\left(Z{m}_{H}+N{m}_{n}-{M}_{\frac{4}{1}x}\right){{c}}^{{2}}$.

## Step 3: 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.

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

Substitute the values in the formula of binding energy.

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

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

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

The error in percentage is given as:

$\begin{array}{l}\text{percentage error}=\frac{\text{true binding energy}}{\text{binding energy}}×100%\\ \text{percentage error}=\frac{51.9\text{eV}}{2.84×{10}^{7}\text{eV}}×100%\\ \text{percentage error}=0.000183%\end{array}$ ### Want to see more solutions like these? 