## Atomic Mass

One unified atomic mass unit (u) is one-twelfth the mass of the carbon-12 atom. \begin{aligned} \, u & = \frac{1}{12} \left( \frac{mass \, of \, 1 \, mole \, of \, carbon \, – 12}{avogadro \, ‘s \, number} \right) \\ & = \frac{1}{12} \left( \frac{0.012 \, kg}{6.02 \times 10^{23}} \right) \\ & = 1.6605 \times 10^{-27} \, kg \end{aligned} The …

## Boltzmann Distribution Function

The total internal energy of an object is the sum of the average energy of each of the atoms that make up the object. Hence, at certain temperature , it is possible that some atoms are more energetic than others.(These atoms occupy a high energy level than others) To find the ratio of number of atoms at a higher energy …

Categories H3

## Derivation Of Compton Shift Equation

Using the above diagram, derive the Compton Shift Equation. Conservation of momentum gives: \begin{aligned}p \, &= p’ \, \text{cos} \, \theta + P_{e} \, \text{cos} \, \phi \\ P_{e} \, \text{cos} \, \phi \, &= p \, – p’ \, \text{cos} \, \theta \end{aligned} AND \begin{aligned} 0 \, &= p’ \, \text{sin} \, \theta \, – P_{e} \, \text{sin} …

Categories H3

## Compton Shift

Using a beam of molybdenum Kɑ X-ray of wavelength 0.0709 nm on a graphite target. There are two peaks in the intensity of the scattered X-rays as a function of the wavelength. The difference between the two peak wavelengths is the Compton shift. The second peak cannot be explained by classical physics. Solved by assuming that the incoming X-ray beam …

Categories H3

## Planck’s Hypothesis

Assumptions used: 1. The particles/oscillators near the surface of the blackbody which emits the blackbody radiation can only have discrete values of energy, En: En=nhf, where n is a positive interger, f is the frequency of the oscillating particle, h is the Planck’s constant. Particles can only have discrete values of energy. Each energy value corresponds to a quantum state …

Categories H3

## Failure Of Classical Theory

Classical View Radiation is wavelike in nature When radiation is incident on an object, the particles near the surface will absorb the radiation and becomes “excited” The excited particles will emit radiation to lose the gained energy and regain stability   Using the above classical view, physicists attempted to explain blackbody radiation. One of the calculations yield Rayleigh-Jeans Law: \$I …

Categories H3