An alpha nucleus of energy 1 2 m v^ 2 bombards a heavy nuclear target of change Ze. Then…
Physics · JEE Main · NTA Exams — Atoms and Nuclei
An alpha nucleus of energy \(\frac{1}{2} m v^{2}\) bombards a heavy nuclear target of change Ze. Then the distance of closest approach for the alpha nucleus will be proportional to
- v2
- 1/m
- \(\frac{1}{v^{4}}\)
- \(\frac{1}{\mathrm{Ze}}\)
Answer
(B) 1/m
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- A radioactive isotope has a half life T years. How long will it take the activity to reduce to 1% of its…
- The volume occupied by an atom is greater than the volume of the nucleus by factor of about:
- The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential…
- In the nuclear reaction, _ 1 ^ 2 H + _ 1 ^ 2 H 1 2 He + 1 n n , if the binding energy of deuteron is 2.23 MeV…
- The relationship between energy of a photon and its wavelength
- C 14 has half life 5700 years. At the end of 11400 years, the actual amount left is
- Let v 1 be the frequency of the series limit of the Lyman series, v 2 be the frequency of the first line of…
- In the case of radio isotope the value of t 1 /2 and λ are identical in magnitude. The value is:
More Atoms and Nuclei questions · Browse all practice questions