In the alpha particle scattering experiment, the distance of closest approach is calculated using the relationship between the initial velocity of the particle and the distance \( d \) between the centers of the alpha particle and the nucleus:
\( r = \frac{d}{2} \)
This formula shows that the distance of closest approach is half the distance between the centers of the two particles when considering the initial velocity of the alpha particle.
Given below are two statements: one is labelled as Assertion (A) and the other is labelled as Reason (R).
Assertion (A): The density of the copper ($^{64}Cu$) nucleus is greater than that of the carbon ($^{12}C$) nucleus.
Reason (R): The nucleus of mass number A has a radius proportional to $A^{1/3}$.
In the light of the above statements, choose the most appropriate answer from the options given below:
Match the LIST-I with LIST-II
LIST-I (Type of decay in Radioactivity) | LIST-II (Reason for stability) | ||
---|---|---|---|
A. | Alpha decay | III. | Nucleus is mostly heavier than Pb (Z=82) |
B. | Beta negative decay | IV. | Nucleus has too many neutrons relative to the number of protons |
C. | Gamma decay | I. | Nucleus has excess energy in an excited state |
D. | Positron Emission | II. | Nucleus has too many protons relative to the number of neutrons |
Choose the correct answer from the options given below: