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Found in: Page 1307

### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718

# Find the disintegration energy Q for the decay of ${{\mathbf{}}}^{{\mathbf{49}}}{\mathbit{V}}$by Kelectron capture (see Problem 54). The needed data are ${{\mathbf{m}}}_{{\mathbf{v}}}{\mathbf{=}}{\mathbf{48}}{\mathbf{.}}{\mathbf{94852}}{\mathbf{}}{\mathbf{u}}{\mathbf{,}}{\mathbf{}}{{\mathbf{m}}}_{{\mathbf{n}}}{\mathbf{=}}{\mathbf{48}}{\mathbf{.}}{\mathbf{94787}}{\mathbf{u}}$ and .

The disintegration energy for the decay of vanadium atom is 600 keV .

See the step by step solution

## Step 1: Identification of given data

The mass of the vanadium atom is ${m}_{v}=48.94852\mathrm{u}$

The mass of the Titanium atom is ${\mathrm{m}}_{\mathrm{Ti}}=48.94787\mathrm{u}$

The energy for K electron capture is ${\mathrm{E}}_{\mathrm{k}}=5.47\mathrm{keV}$

## Step 2: Concept Introduction

The disintegration energy for an atom is the energy necessary for breaking the nucleus of the atom.

The disintegration energy for the decay of vanadium atom is given as:

${\mathbf{E}}{\mathbf{=}}\left({m}_{v}-{m}_{\mathrm{Ti}}\right){{\mathbf{c}}}^{{\mathbf{2}}}{\mathbf{-}}{{\mathbf{E}}}_{{\mathbf{k}}}$ …………………….(1)

Here, is the speed of light.

## Step 3: Determination of disintegration energy for the decay of vanadium atom

Substitute all the values in the above equation (1), and we get.

$\mathrm{E}=\left(48.94852\mathrm{u}-48.94787\mathrm{u}\right){\mathrm{c}}^{2}-5.47\mathrm{keV}\phantom{\rule{0ex}{0ex}}=\left(0.00065\mathrm{u}\right)\left({\mathrm{c}}^{2}\right)\left(\left(\frac{1}{{\mathrm{c}}^{2}}\right)\left(\frac{931.5×{10}^{3}\mathrm{keV}}{\mathrm{u}}\right)\right)-5.47\mathrm{keV}\phantom{\rule{0ex}{0ex}}=600\mathrm{keV}$

Therefore, the disintegration energy for the decay of vanadium atom is 100 keV.