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Modern Physics
Found in: Page 518
Modern Physics

Modern Physics

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 24.6eV. 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?

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

See the step by step solution

Step by Step Solution

Step 1: Given data

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

Energy required remove second electron =54.5eV .

Step 2: Formula for Binding energy

For helium- 4, binding energy is given by, BE=(ZmH+NmnM41x)c2.

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:


Simplify further as shown below.

M42x=4.002603uc=3×108 m/s

Substitute the values in the formula of binding energy.

BE=[2(1.007825u)+2(1.008665u)4.002603u(3×108 m/s)2BE=(0.030377u(1.661×1021kg1u))(3×108 m/s)2BE=4.54×1012 J(1eV1.6×1019 J)BE=2.84×107eV

True binding energy for helium nuclei =24.6eV+54.5eV13.6eV×2=51.9eV.

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

The error in percentage is given as:

percentage error = true binding energy binding energy ×100%percentage error=51.9eV2.84×107eV×100%percentage error=0.000183%

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