PhysicsAtomsClass 12

Atoms: Rutherford, Bohr and X-rays: interactive simulation

Rutherford fired α particles at gold foil and found that a few bounced almost straight back. Only a tiny, heavy, positive nucleus could do that — and the size of the nucleus came out of the geometry of the scattering.

Useful for: JEE Main, JEE Advanced, NEET, CBSE Boards, AP Physics 2, A-level Physics, IB Physics · 3 modes in one lab

What you can explore

  1. 1

    Rutherford scattering

    Impact parameter, scattering angle and distance of closest approach for gold, silver and aluminium.
  2. 2

    Bohr orbits

    Radii, speeds and energies for H, He⁺ and Li²⁺ with the standing electron wave.
  3. 3

    X-ray spectrum

    Continuous spectrum and cut-off wavelength, with Kα and Kβ lines for Cu, Mo and W.

Try this

Change the impact parameter and energy of the α particles, step through Bohr’s quantised orbits and watch the electron wave fit around them, then switch on an X-ray tube and see which parts of its spectrum depend on the voltage and which on the target.

  1. 1

    Fire 7.7 MeV α particles head-on (b = 0) at gold.

    What you should see

    They stop at r₀ = 2 × 79 × 1.44 / 7.7 ≈ 29.5 fm and come straight back.

  2. 2

    Set b equal to r₀.

    What you should see

    tan(θ/2) = r₀/2b = 1/2, so θ ≈ 53°.

  3. 3

    Switch the target from gold to aluminium.

    What you should see

    r₀ shrinks by 79/13 and all paths bend much less.

  4. 4

    In Bohr orbits, compare n = 1 of H with n = 1 of He⁺.

    What you should see

    He⁺ has half the radius (0.265 Å) and four times the binding energy (−54.4 eV).

  5. 5

    In X-ray spectrum, use a molybdenum target at 30 kV.

    What you should see

    λ_min ≈ 41.3 pm. The Kα line (≈ 72 pm estimated) appears only above about 23 kV.

Formulas behind it

Distance of closest approach (head-on)

r0=2Ze24πε0Kr_0 = \frac{2Ze^2}{4\pi\varepsilon_0 K}

Impact parameter

b=r02cot⁡θ2b = \frac{r_0}{2}\cot\frac{\theta}{2}

Bohr radius and energy

rn=0.529n2Z A˚,En=−13.6Z2n2 eVr_n = 0.529\frac{n^2}{Z}\ \text{Å},\qquad E_n = -13.6\frac{Z^2}{n^2}\ \text{eV}

Quantisation condition

mvr=nh2π  ⟺  2πr=nλmvr = \frac{nh}{2\pi} \iff 2\pi r = n\lambda

X-ray cut-off wavelength

λmin=hceV\lambda_{min} = \frac{hc}{eV}

Moseley’s law

ν=a(Z−b)\sqrt{\nu} = a(Z - b)

Common mistakes

  • Forgetting the factor 2 for the α particle’s charge in the closest-approach formula.
  • Using the electron’s mass in the Bohr formulas for He⁺ but not changing Z.
  • Thinking λ_min depends on the target metal. It depends only on the accelerating voltage.
  • Mixing up continuous and characteristic X-rays: only the characteristic lines identify the element.
Practise 239 questionsCheck your answer instantly, with a worked solution.Modern Physics chapter guideFormula sheet, solved examples and PYQ patterns.Stuck on a atoms: rutherford, bohr and x-rays question?Snap it and get a whiteboard video solution.

FAQ

Why did most α particles pass straight through the foil?

The atom is mostly empty space. Only particles passing within a few tens of femtometres of a nucleus feel a large enough force to be deflected noticeably.

Why does the Bohr model fail for helium?

It handles only one electron. With two or more, the electrons repel each other and the simple orbit picture breaks down.

What did Moseley’s law show?

That the square root of the X-ray frequency increases in a straight line with atomic number, so elements should be ordered by Z, not atomic mass.

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