Book edition 2nd Edition

Author(s) Randy Harris

Pages 633 pages

ISBN 9780805303087

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Short Answer

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 $0.000183 %$ .

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Energy required removing one electron from helium atom $m = 24.6 eV$ .

Energy required remove second electron $= 54.5 eV$ .

Step 2: Formula for Binding energy

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

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:

$\begin{matrix} Z = 2 \\ m_{H} = 1.007825 u \\ N = 2 \\ m_{n} = 1.008665 u \end{matrix}$

Simplify further as shown below.

$\begin{matrix} M_{\frac{4}{2} x} = 4.002603 u \\ c = 3 \times 10^{8} m / s \end{matrix}$

Substitute the values in the formula of binding energy.

$\begin{array}{l} B E = [2 (1.007825 u) + 2 (1.008665 u) - 4.002603 u ∣ {(3 \times 10^{8} m / s)}^{2} \\ B E = (0.030377 u (\frac{1.661 \times 10^{- 21} k g}{1 u})) {(3 \times 10^{8} m / s)}^{2} \\ B E = 4.54 \times 10^{- 12} J (\frac{1 eV}{1.6 \times 10^{- 19} J}) \\ B E = 2.84 \times 10^{7} eV \end{array}$

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

Therefore true binding energy of helium nuclei should be $51.9 eV$ higher.

The error in percentage is given as:

\begin{array}{l} percentage error = \frac{true binding energy}{binding energy} \times 100 % \\ percentage error = \frac{51.9 eV}{2.84 \times 10^{7} eV} \times 100 % \\ percentage error = 0.000183 % \end{array}

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Modern Physics

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Short Answer

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?

Step by Step Solution

Step 1: Given data

Step 2: Formula for Binding energy

Step 3: Calculation for the error in the binding energy of helium nuclei

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Modern Physics

Answers without the blur.

Short Answer

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?

Step by Step Solution

Step 1: Given data

Step 2: Formula for Binding energy

Step 3: Calculation for the error in the binding energy of helium nuclei

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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?