Question

Find the disintegration energy Q for the decay of ${\mathbf{}}^{\mathbf{49}}\mathit{V}$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 .

Short answer

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

Step-by-step solution

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 ${\mathrm{E}}_{\mathrm{k}}=5.47\mathrm{keV}$

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.

Determination of disintegration energy for the decay of vanadium atom

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

$$\begin{gathered} \mathrm{E}=(48.94852\mathrm{u}-48.94787\mathrm{u}){\mathrm{c}}^2-5.47\mathrm{keV} \\ =\left(0.00065\mathrm{u}\right)\left({\mathrm{c}}^2\right)\left(\left(\frac{1}{{\mathrm{c}}^2}\right)\left(\frac{931.5\times {10}^3\mathrm{keV}}{\mathrm{u}}\right)\right)-5.47\mathrm{keV} \\ =600\mathrm{keV} \end{gathered}$$

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