1.7MeV金原子撞击镁原子：弹性碰撞后镁原子能量范围求解

Hey there! Let's walk through how to calculate the possible energy range of the magnesium atom after this collision. We'll stick to elastic collision assumptions here (since you didn't mention any inelastic processes like nuclear excitation or electron shell ionization—those would complicate things, but we can cover the standard case first).

Step 1: Grab the atomic masses #

First, we need the relative masses of the two atoms (using atomic mass units, u):

• Gold (Au): ( M = 197 , \text{u} )
• Magnesium (Mg): ( m = 24 , \text{u} )
  The incoming gold atom has an initial energy ( E_0 = 1.7 , \text{MeV} ).

Step 2: Maximum energy transfer (head-on collision) #

The magnesium atom will get the most energy when the gold atom slams directly into it (a head-on, central collision). For elastic collisions, the formula for maximum recoil energy of the target atom is:

E_max = E₀ * (4*M*m) / (M + m)²

Plugging in the numbers:

• Numerator: ( 4 * 197 * 24 = 18912 )
• Denominator: ( (197 + 24)² = 221² = 48841 )
• The ratio is ~0.387, so ( E_max ≈ 1.7 * 0.387 ≈ 0.66 , \text{MeV} )

Step 3: Minimum energy transfer (grazing collision) #

At the other extreme, if the gold atom just barely "brushes" the magnesium atom (a grazing collision, where the impact parameter is huge), the magnesium atom gets almost no momentum transfer. Its recoil energy will be practically 0—we can say the minimum energy approaches 0.

Final Energy Range #

Putting it all together, the possible energy of the magnesium atom after collision is:

( 0 < E_{\text{Mg}} ≤ 0.66 , \text{MeV} )

Quick Extra Note #

If you wanted to calculate the energy for any scattering angle, there's a general formula:

E(θ) = E₀ * (4*M*m)/(M + m)² * cos²φ

Here, φ is the recoil angle of the magnesium atom. Since ( cos²φ ) ranges from 0 to 1, this directly confirms our range—when φ=0 (head-on), we get the max energy; when φ approaches 90°, energy drops to near 0.

内容的提问来源于stack exchange，提问作者Jackson Hart