Question

A nucleus \({ }_{n} X^{\mathrm{m}}\) emist one \(\alpha\) - Particle and two \(\beta\) -Particle. The resulting nucleus is (A) \(_{n-2} Y^{m-4}\) (B) \(_{\mathrm{n}} \mathrm{Y}^{\mathrm{m}-6}\) (C) \(\mathrm{n} \mathrm{Y}^{\mathrm{m}-4}\) (D) \(_{n-4} Y^{m-6}\)

Short answer

The resulting nucleus after emitting one alpha particle and two beta particles is \(_{\mathrm{n}} \mathrm{Y}^{\mathrm{m}-4}\), which corresponds to option (C).

Step-by-step solution

Emission of an alpha particle

When the nucleus emits an alpha particle, it loses 2 protons and 2 neutrons. So, the mass number (m) decreases by 4 and the atomic number (n) decreases by 2. After the emission of an alpha particle, the resulting nucleus is: \(_{n-2}X^{m-4}\)

First beta particle emission

In beta particle emission, if a neutron decays into a proton, the mass number remains the same while the atomic number increases by 1. After the first beta particle emission, the resulting nucleus becomes: \(_{(n-2)+1}X^{m-4} = _{n-1}X^{m-4}\)

Second beta particle emission

Similarly, for a second beta particle emission, we get: After the second beta particle emission, the resulting nucleus becomes: \(_{(n-1)+1}X^{m-4} = _{n}X^{m-4}\) However, since the nucleus has changed due to these emissions, we can't use the symbol X. Instead, we use the symbol Y to represent the new nucleus. Therefore, the resulting nucleus is: \(_{\mathrm{n}} \mathrm{Y}^{\mathrm{m}-4}\) Comparing our result with the given options, we find that our answer corresponds to option (C).