The minimum frequency of photon required to break a particle of mass 15.348 amu into 4alpha$\alpha$ particles is \_\_\_\_ kHz. [mass of He nucleus = 4.002 amu, 1 amu = 1.66 times 10^-27$1.66 \times 10^{-27}$ kg, h = 6.6 times 10^-34$6.6 \times 10^{-34}$ J.s and c = 3 times 10^8$3 \times 10^{8}$ m/s]

### Related Formula r_0 = frac14piepsilon_0 fracq_1 q_2K$$r_0 = \frac{1}{4\pi\epsilon_0} \frac{q_1 q_2}{K}$$ ### Core Logic

Using mechanical energy conservation: PE_i + KE_i = PE_f + KE_f$$PE_i + KE_i = PE_f + KE_f$$ 0 + 7.9 times 10^6 times 1.6 times 10^-19 = frack(2e)(Ze)r + 0$$0 + 7.9 \times 10^6 \times 1.6 \times 10^{-19} = \frac{k(2e)(Ze)}{r} + 0$$ r = frac9 times 10^9 times 2 times (1.6 times 10^-19)^2 times 797.9 times 10^6 times 1.6 times 10^-19 = 2.88 times 10^-14 text m$$r = \frac{9 \times 10^9 \times 2 \times (1.6 \times 10^{-19})^2 \times 79}{7.9 \times 10^6 \times 1.6 \times 10^{-19}} = 2.88 \times 10^{-14} \text{ m}$$ For diameter: D = 2r = 5.76 times 10^-14 text m$$D = 2r = 5.76 \times 10^{-14} \text{ m}$$ ### Pattern Recognition Sees: Alpha particle scattering + atomic number + kinetic energy. Shortcut: Equate initial kinetic energy to electrostatic potential energy at closest approach distance r$r$, then double for diameter D = 2r$D = 2r$. Check: Results align with option (1). ✓ ### Chapter Mix Class 12 Physics: Atoms and Nuclei