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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

Complete the reaction ${ }_{0} \mathrm{n}^{1}+{ }_{92} \mathrm{U}^{235} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}+3\left({ }_{0} \mathrm{n}^{1}\right)$ (A) \(_{36} \mathrm{Kr}^{90}\) (B) \(_{36} \mathrm{Kr}^{89}\) (C) \(_{36} \mathrm{Kr}^{91}\) (D) \(_{36} \mathrm{Kr}^{92}\)

The radius of Ge nucleus is measured to be twice the radius of ${ }^{9}{ }_{4}$ Be. The number of nucleons in Ge are (A) 72 (B) 78 (C) 65 (D) 80

if \(_{92} \mathrm{U}^{238}\) undergoes successively \(8 \alpha\) decays and $6 \beta$ decays then resulting nucleus is (A) \(\mathrm{Pb}^{206}\) (B) \(\mathrm{Pb}^{208}\) (C) \(\mathrm{Pb}^{214}\) (D) None of these

when \(u-238\) nucleus originally at rest decay by emitting an \(\alpha\) -particle having a speed u the recoil speed of the residual nucleus is. (A) \([(4 \mathrm{u}) /(238)]\) (B) \([(-4 \mathrm{u}) /(238)]\) (C) \([(4 \mathrm{u}) /(234)]\) (D) \([(-4 \mathrm{u}) /(234)]\)

The wave length of the first line of Lyman series for hydrogen atom is equal to that of hydrogen atom is equal to that of second line of Balmar series for a hydrogen like ion. The atomic number \(\mathrm{Z}\) of hydrogen like ion is (A) 1 (B) 2 (C) 3 (D) 4
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