Skip to content
Question of 50

Q.Half life of U-238 undergoing α - decay is 4.5 × 10⁹ years. What is the activity of one gram of U-238 sample?

Karnataka PUCKarnataka II PUC Board 2019Subjective· 5mImportance★★★★★
0% · 0/50 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Find λ=0.693/T1/2\lambda = 0.693/T_{1/2} and NN = atoms in 1 g, then A=λN≈1.23×104 BqA = \lambda N \approx 1.23\times10^{4}\ \text{Bq}.

Given. T1/2=4.5×109 yearsT_{1/2} = 4.5\times10^{9}\ \text{years}; mass m=1 gm = 1\ \text{g} of 238U^{238}\text{U} (molar mass 238 g mol−1238\ \text{g mol}^{-1}).

Decay constant. Converting the half-life to seconds (1 year=3.156×107 s1\ \text{year} = 3.156\times10^{7}\ \text{s}):

T1/2=4.5×109×3.156×107=1.42×1017 s.T_{1/2} = 4.5\times10^{9}\times3.156\times10^{7} = 1.42\times10^{17}\ \text{s}.

λ=0.693T1/2=0.6931.42×1017=4.88×10−18 s−1.\lambda = \frac{0.693}{T_{1/2}} = \frac{0.693}{1.42\times10^{17}} = 4.88\times10^{-18}\ \text{s}^{-1}.

Number of atoms in 1 g.

N=mM NA=1238×6.022×1023=2.53×1021 atoms.N = \frac{m}{M}\,N_A = \frac{1}{238}\times6.022\times10^{23} = 2.53\times10^{21}\ \text{atoms}.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.