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JEE Mains · Physics · STD 12 - 13. Nuclei

\(N_{\beta}\) એ \(1\) ગ્રામ \(Na^{24}\) ના રેડિયોએક્ટિવ ન્યુક્લિયસમાંથી(અર્ધઆયુષ્ય સમય\(= 15\, hrs\)) \(7.5\, hours\) માં ઉત્સર્જિત થતાં \(\beta\) કણોની સંખ્યા છે તો \(N_{\beta}\) નું મૂલ્ય લગભગ કેટલું હશે? (એવોગેડ્રો નંબર\(= 6.023\times10^{23}\,/g.\, mole\))

  1. A \(6.2\times10^{21}\)
  2. B \(7.5\times10^{21}\)
  3. C \(1.25\times10^{22}\)
  4. D \(1.75\times10^{22}\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(7.5\times10^{21}\)

Step-by-step Solution

Detailed explanation

We know that \(N_{\beta}=N_{0}\left(1-e^{-\lambda t}\right)\) \(N_{\beta}=\frac{6.023 \times 10^{23}}{24} \cdot\left[1-e^{\left(-\frac{\ln ^{2} }{15} \times 75\right)}\right]\) on solving we get, \(\mathrm{N}_{\beta}=7.4 \times 10^{21}\)
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