JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
An \(\alpha\) particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particals will be.
- A \(\sqrt{2}: 1\)
- B \(2 \sqrt{2}: 1\)
- C \(4 \sqrt{2}: 1\)
- D \(8: 1\)
Answer & Solution
Correct Answer
(B) \(2 \sqrt{2}: 1\)
Step-by-step Solution
Detailed explanation
\(p =\sqrt{2 mE }=\sqrt{2 mqV }\) \(\frac{ p _{\alpha}}{ p _{ p }}=\sqrt{\frac{ m _{\alpha} q _{\alpha}}{ m _{ p } q _{ p }}}=\sqrt{\frac{4}{1} \times \frac{2}{1}}\) \(=\frac{2 \sqrt{2}}{1}\)
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