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(a) ʃ [√expiФ]2dФ= ʃ, assuming spherical nature of the light atom, we use spherical polar coordinates, x=rsinөcosҨ, y=rsinөsinҨ, z=rcosө, dФ=r2sinөdrdөdҨ, where r= 0 to∞, ө= 0 to ∏, Ҩ= 0 to ∏,
6. for harmonic oscillator, zero point energy = potential energy, given by V(x)=0.5kf x2. From relation F=ke and that F=ma, But each mass has extension of = 0.03448m, hence for combination of two masses, total extension= 0.03445/2=0.01724m. here the gravitational intensity is assumed to be 10N/Kg. hence V(x)= 0.5×725×0.0003= 0.10875J. Comparing with thermal energy at 298K, i.e 4.11×10-21J, the zero point energy is greater showing that the population of particles in the energy level with similar amount of energy as zero point energy is very minimal.
8. (a) Assume the potential for bond breaking is harmonic, then V(x)= 0.5kfx2 where kf= force constant and x is extension of the bond from equilibrium position. Therefore, 0.5kfx2= Vm(x), thus 0.5kfx2= De[1-exp(-ax)]2, but De=7.70×10-19J, kf=412N/M, therefore kf= which can be expressed as ...Show more

Summary

= sinө[cosөd/dөsinө+sinөd/dөcosө]{sinө[cosөd/dөsinө+sinөd/dөcosө]}+6sin2ө =sinө[cos2ө-sin2ө]{sinө[cos2ө-sin2ө]}+6sin2ө, from this equation it is clear that the function cannot be regenerated.
4. for the particle moving along a circle covers some radius r…
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