To solve this problem, we need to analyze the situation involving the nucleus X, the emitted alpha particle, and the resulting nucleus Y. Let's break down the information given and apply some nuclear physics concepts to find the value of n.
Understanding the Components
We have a nucleus X with a mass number that we can denote as A_X. When it emits an alpha particle, which consists of 2 protons and 2 neutrons, it transforms into nucleus Y. The kinetic energy of the emitted alpha particle is given as 9.0 MeV.
Binding Energy Concepts
Binding energy per nucleon is a crucial concept in nuclear physics. It indicates how tightly the nucleons (protons and neutrons) are bound within a nucleus. The higher the binding energy per nucleon, the more stable the nucleus is. In this problem, we know:
- The binding energy per nucleon of the alpha particle is different from that of nucleus Y by 1.5 MeV.
- The difference in binding energy per nucleon between nuclei X and Y is given as (n/230) MeV.
Calculating the Binding Energies
Let’s denote the binding energy per nucleon of the alpha particle as BE_alpha, and that of nucleus Y as BE_Y. According to the problem, we have:
- BE_Y = BE_alpha + 1.5 MeV
Now, for nucleus X, we denote its binding energy per nucleon as BE_X. The relationship given in the problem states:
Substituting Values
We can substitute the expression for BE_Y into the equation involving BE_X:
- BE_X - (BE_alpha + 1.5) = n/230
Rearranging this gives us:
- BE_X - BE_alpha - 1.5 = n/230
Finding the Binding Energy Difference
Now, we need to express BE_X in terms of BE_alpha. The binding energy per nucleon of the alpha particle is typically around 7.1 MeV. If we assume this value, we can calculate:
- BE_X = BE_alpha + (n/230) + 1.5
Substituting BE_alpha = 7.1 MeV:
- BE_X = 7.1 + (n/230) + 1.5
- BE_X = 8.6 + (n/230)
Setting Up the Equation
Now, we can analyze the difference between BE_X and BE_Y:
- BE_X - BE_Y = (8.6 + (n/230)) - (7.1 + 1.5)
- BE_X - BE_Y = (8.6 - 8.6) + (n/230) = n/230
Since we know that BE_X - BE_Y = n/230, we can conclude that:
Solving for n
From the equation above, we can see that for the difference in binding energies to hold true, n must equal 0. Thus, the value of n is:
This means that the binding energy per nucleon of nucleus X and nucleus Y are equal, indicating that the emission of the alpha particle did not change the overall binding energy per nucleon significantly in this scenario. If you have any further questions or need clarification on any part of this explanation, feel free to ask!
