Question

The uranium nucleus $_{92}{U^{238}}$ emits a $\alpha$−particle and the resulting nucleus emits one $\beta$ −particle. The atomic number and mass number of the final nucleus will be respectively
A) 91,234
B) 90,234
C) 91,238
D) 92,234

Answer 1

It is clear that in decay the proton stays in the nucleus but the electron leaves the atom as a beta particle. In alpha decay mass and atomic both will change by subtraction of 4 and 2 whereas in beta decay only mass number increases.

Step by step solution:
Step 1:
The emission of alpha particles is a property of the heaviest nuclei, such as uranium 238 with its 92 protons and 136 neutrons, the heaviest natural nucleus observed. These unstable nuclei emit a light helium nucleus in order to reduce their mass and hence increase their stability.
The proton stays in the nucleus but the electron leaves the atom as a beta particle. When a nucleus emits a beta particle, these changes happen: the mass number stays the same. The atomic number increases by 1.

Step 2:
For more clarity have a look on the diagram:

Release of a α will decrease the atomic number by 2 and mass number by 4, and β particle will increase the mass number by 1 without change in mass number.
So final product will have a mass number of 234 and atomic number =91.

Hence option A is correct.

Additional information:
Alpha particles are dangerous: alpha radiation is the most dangerous because it is easily absorbed by cells. Beta and gamma radiation are not as dangerous because they are less likely to be absorbed by a cell and will usually just pass right through it.

Note: All natural uranium isotopes emit alpha particles – positively charged ions identical to the nucleus of a helium atom, with two protons and two neutrons. Their beta and gamma activity is low. Alpha particles are relatively large, and do not penetrate far in tissue – they are stopped by the skin, for example.