Question

If \(S=\{1,2,....,50\}\), two numbers \(\alpha\) and \(\beta\) are selected at random find the probability that product is divisible by 3 :

• A. \(\frac{664}{1225}\)
• B. \(\frac{646}{1225}\)
• C. \(\frac{527}{1225}\)
• D. \(\frac{461}{1225}\)

Hint:

For probability problems involving conditions like "at least one" or "divisible by," it is often much simpler to calculate the probability of the complementary event ("none" or "not divisible by") and subtract it from 1.

Answer 1

The Correct Option is A

Concept: This is a photon-induced nuclear reaction. The photon must supply enough energy to overcome the mass defect (binding energy difference) between the initial nucleus and the products (4 alpha particles). \[ E_{photon} = \Delta m c^2 = h\nu \] Step 1: Calculate Mass Defect. Initial mass: \(15.356\,\text{amu}\). Final mass (4 \(\alpha\)): \(4 \times 4.004 = 16.016\,\text{amu}\). Note that the products are heavier than the reactant, meaning energy must be supplied. \[ \Delta m = 16.016 - 15.356 = 0.660\,\text{amu} \]
Step 2: Convert Mass to Energy. Using \(1\,\text{amu} \approx 1.66 \times 10^{-27}\,\text{kg}\): \[ \Delta E = \Delta m c^2 = (0.660 \times 1.66 \times 10^{-27}) \times (3 \times 10^8)^2 \] \[ \Delta E \approx 1.0956 \times 10^{-27} \times 9 \times 10^{16} \approx 9.86 \times 10^{-11}\,\text{J} \]
Step 3: Calculate Frequency. \[ \nu = \frac{\Delta E}{h} = \frac{9.86 \times 10^{-11}}{6.6 \times 10^{-34}} \approx 1.49 \times 10^{23}\,\text{Hz} \]
Step 4: Convert unit to kHz. \[ \nu = 1.49 \times 10^{20}\,\text{kHz} \approx 14.9 \times 10^{19}\,\text{kHz} \] \[ \boxed{\nu = 14.9\times10^{19}\,\text{kHz}} \]