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What is induced radioactivity? Name the radioactive element obtained when $ _{13}A{l^{27}} $ is bombarded with alpha particles. Write one use of this radioactive element.

Answer
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Hint: To answer this question, we have to look into the basic definition of the induced radioactivity. Then, we have to write a nuclear reaction corresponding to the situation given in the question, with the help of which we will be able to predict the radioactive element obtained.

Complete step by step answer
We know that there are some nuclei which are not stable under normal conditions, and therefore they get disintegrated into stable nuclei. This is known as radioactivity. The elements, which are capable of showing radioactivity, are termed as radioactive elements. The examples of such elements include uranium, radium, thorium etc. This phenomenon was discovered by A.H. Becquerel.

But in 1934, Irene-Joliot-Curie with Frederic-Joliot-Curie discovered the phenomenon of induced radioactivity, where a stable nucleus can be made radioactive by using radiation. It happens in the case of the lighter elements, such as Boron and Aluminium. When the alpha particles are bombarded onto them, the lighter particles emit radiation. This emission is continued even after stopping the alpha radiations. Thus, the lighter element turns into a radioactive element. The radiation which is emitted consists of positrons, which has a mass equal to the electron, but a charge opposite to it.

When an aluminium nucleus is bombarded with an alpha particle (a particle similar to the helium nucleus) the following reaction takes place.
 $ _{13}A{l^{27}}{ + _2}H{e^4}{ \to _{15}}{P^{30}}{ + _0}{n^1} $

The element $ _{15}{P^{30}} $ obtained in the above reaction is called the radio phosphorous.
It is used in radiotherapy for the treatment of blood disorders.

Note
Whenever we write a nuclear reaction, we have to take proper care of the mass and the charge conservation. The mass is represented by the mass number of the elements present, while the charge is represented by their atomic numbers.