Find the disintegration energy Q for the decay of 49Vby Kelectron capture (see Problem 54). The needed data are mv=48.94852u,mn=48.94787uand .

Short Answer

Expert verified

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

Step by step solution

01

Identification of given data

The mass of the vanadium atom is mv=48.94852u

The mass of the Titanium atom is mTi=48.94787u

The energy for K electron capture is Ek=5.47keV

02

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:

E=(mv-mTi)c2-Ek…………………….(1)

Here, is the speed of light.

03

Determination of disintegration energy for the decay of vanadium atom

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

E=(48.94852u-48.94787u)c2-5.47keV=0.00065u(c2)1c2931.5×103keVu-5.47keV=600keV

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

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