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Chapter 18: Problem 2555

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

Expert verified
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

01

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: \(_{n-2}X^{m-4}\)
02

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: \(_{(n-2)+1}X^{m-4} = _{n-1}X^{m-4}\)
03

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).

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Most popular questions from this chapter

In terms of Rydberg constant \(R\). The wave number of first Balmer line is (A) \((5 \mathrm{R} / 36)\) (B) \((8 \mathrm{R} / 9)\) (C) \(\mathrm{R}\) (D) \((8 \mathrm{R} / 20)\)

If a hydrogen atom emits a Photon of wave length \(\lambda\). the recoil speed of the atom of mass \(\mathrm{m}\) is given by (A) \((\mathrm{h} / \mathrm{m} \lambda)\) (B) \((\mathrm{mh} / \lambda)\) (C) \(\operatorname{mh} \lambda\) (D) \((\mathrm{m} \lambda / \mathrm{h})\)

Match column I and II and chose correct Answer form the given below. (a) Nuclear fusion (p) converts some matter into energy (b) Nuclear fission (q) generally Possible for nuclei with low atomic number (c) \(\beta\) decay (r) generally Possible for nuclei with high atomic number (d) Exothermic nuclear (s) Essentially Proceeds by weak reaction nuclear force(c) (A) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (B) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{p}$ (D) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{q}, \mathrm{c} \rightarrow \mathrm{p}, \mathrm{d} \rightarrow \mathrm{s}$

\({ }_{6} \mathrm{C}^{12}\) absorbs an energetic neutron and emits a \(\beta\) Particle. The resulting nucleus is (A) \({ }_{7} \mathrm{~N}^{13}\) (B) \({ }_{7} \mathrm{~N}^{14}\) (C) \({ }_{6} \mathrm{C}^{13}\) (D) \({ }_{6} \mathrm{C}^{12}\)

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