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JEE Mains · Physics · STD 12 - 13. Nuclei

\({ }_{92}^{238} A \rightarrow{ }_{90}^{234} B +{ }_2^4 D + Q\) આપેલ ન્યૂક્લિયર પ્રક્રિયામાં, મુક્ત થતી ઊર્જાનું અંદાજિત (સંનિકટ) મૂલ્ય \(..........\,MeV\) હશે. \({ }_{92}^{238} A=\) નું દળ \(238.05079 \times 931.5\,MeV / c ^2\) \({ }_{90}^{234} B =\) નું દળ \(234.04363 \times 931.5\,MeV / c ^2\) \({ }_2^4 D =\) નું દળ \(4.00260 \times 931.5\,MeV / c ^2\) આપેલ છે.

  1. A \(3.82\)
  2. B \(5.9\)
  3. C \(2.12\)
  4. D \(4.25\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(4.25\)

Step-by-step Solution

Detailed explanation

\(=\left(m_A-m_B-m_D\right) \times 931.5\,MeV\) \(=(238.05079-234.04363-4.00260) \times 931.5\) \(\Rightarrow 4.25\,Mev\)
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