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NEET PHYSICSATOMSMedium

Question

If the nucleus 1327Al^{27}_{13}\mathrm{Al} has a nuclear radius of about 3.6 fermis, then 52125Te^{125}_{52}\mathrm{Te} would have its radius approximately as:

A

6.0 Fermi

B

9.6 Fermi

C

12.0 Fermi

D

4.8 Fermi

Step-by-Step Solution

The nuclear radius RR is related to the mass number AA by the relationship R=R0A1/3R = R_0 A^{1/3}, where R0R_0 is a constant (approximately 1.2 fm).

For Aluminum (AlAl), the mass number A1=27A_1 = 27 and radius R1=3.6R_1 = 3.6 fm. For Tellurium (TeTe), the mass number A2=125A_2 = 125.

Taking the ratio: R2R1=(A2A1)1/3\frac{R_2}{R_1} = \left(\frac{A_2}{A_1}\right)^{1/3} R23.6=(12527)1/3\frac{R_2}{3.6} = \left(\frac{125}{27}\right)^{1/3} R23.6=53\frac{R_2}{3.6} = \frac{5}{3} R2=3.6×53=1.2×5=6.0R_2 = 3.6 \times \frac{5}{3} = 1.2 \times 5 = 6.0 fm.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from ATOMS. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSATOMSnucleusmathrmalnuclearradiusfermis

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