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Chapter 19: Problem 7

An isotope of rutherfordium, \({ }_{10}^{257} \mathrm{Rf}\), is formed by the bombardment of californium- 249 by carbon-12. In the process, neutrons are emitted. The new isotope formed decays rapidly, emitting an alpha particle. (a) How many neutrons are emitted for every Cf-249 bombarded? (b) Write the nuclear symbol for the isotope formed by the decay of Rf-257.

Short Answer

Expert verified
4 neutrons are emitted for every Californium-249 bombarded. (b) What is the nuclear symbol for the isotope formed by the decay of Rutherfordium-257? The nuclear symbol for the isotope formed by the decay of Rutherfordium-257 is ${}_{102}^{253}\mathrm{X}$.

Step by step solution

01

Identify the conservation of nucleons in a nuclear reaction

In a nuclear reaction, the total number of nucleons (protons + neutrons) remains constant. This means, the sum of the nucleons before the reaction equals the sum of nucleons after the reaction.
02

Write the given nuclear reaction

In this problem, Californium-249 (\({ }_{98}^{249}\mathrm{Cf}\)) is bombarded by Carbon-12 (\({ }_{6}^{12}\mathrm{C}\)) forming an isotope of Rutherfordium (\({ }_{10}^{257}\mathrm{Rf}\)) and emitting neutrons (\(n\)). The nuclear equation for this reaction is: \({ }_{98}^{249}\mathrm{Cf} + { }_{6}^{12}\mathrm{C} \rightarrow { }_{104}^{257}\mathrm{Rf} + x\cdot n\) where `x` is the number of neutrons emitted during the reaction.
03

Use conservation of nucleons to find the number of emitted neutrons

According to the conservation of nucleons, the total number of nucleons before the reaction equals the total number of nucleons after the reaction: 249 + 12 = 257 + x Now, we solve for x to find the number of neutrons emitted: x = 261 - 257 = 4 Thus, 4 neutrons are emitted for every Californium-249 bombarded. #Part (b): Writing the nuclear symbol for the isotope formed by the decay of Rutherfordium-257#
04

Identify the process of alpha decay

Alpha decay is a process in which an unstable nucleus emits an alpha particle, leading to a decrease in the atomic number by 2 and the mass number by 4. An alpha particle consists of 2 protons and 2 neutrons (\({ }_{2}^{4}\mathrm{He}\)).
05

Write the alpha decay nuclear equation for Rf-257

The alpha decay of Rutherfordium-257 can be represented as follows: \({ }_{104}^{257}\mathrm{Rf} \rightarrow { }_{Z}^{A}\mathrm{X} + { }_{2}^{4}\mathrm{He}\) where `X` is the nucleus formed after the decay, `Z` is its atomic number, and `A` is its mass number.
06

Calculate the new atomic number and mass number

To find the new atomic number and mass number, we use the conservation of nucleons. The sum of nucleons before the decay should be equal to the sum of nucleons after the decay. For the atomic numbers, we have: 104 = Z + 2 So, the new atomic number Z is 102. For the mass numbers, we have: 257 = A + 4 So, the new mass number A is 253. Thus, the nuclear symbol for the isotope formed by the decay of Rutherfordium-257 is \({ }_{102}^{253}\mathrm{X}\).

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