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lianna [129]
3 years ago
15

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,

Physics
1 answer:
madam [21]3 years ago
5 0

Answer:

  • <u><em>(b) 1:1</em></u>

Explanation:

<u></u>

<u>1. Formulae:</u>

  • E = hf  
  • E = h.v/λ
  • λ = h/(mv)
  • E = (1/2)mv²

Where:

  • E = kinetic energy of the particle
  • λ = de-Broglie wavelength
  • m = mass of the particle
  • v = speed of the particle
  • h = Planck constant

<u><em>2. Reasoning</em></u>

An alha particle contains 2 neutrons and 2 protons, thus its mass number is 4.

A proton has mass number 1.

Thus, the relative masses of an alpha particle and a proton are:

       \dfrac{m_\alpha}{m_p}=4

For the kinetic energies you find:

          \dfrac{E_\alpha}{E_p}=\dfrac{m_\alpha \times v_\alpha^2}{m_p\times v_p^2}

            \dfrac{1eV}{4eV}=\dfrac{4\times v_\alpha^2}{1\times v_p^2}\\\\\\\dfrac{v_p^2}{v_\alpha^2}=16\\\\\\\dfrac{v_p}{v_\alpha}=4

Thus:

           \dfrac{m_\alpha}{m_p}=4=\dfrac{v_p}{v__\alpha}

          m_\alpha v_\alpha=m_pv_p

From de-Broglie equation, λ = h/(mv)  

       \dfrac{\lambda_p}{\lambda_\alpha}=\dfrac{m_\lambda v_\lambda}{m_pv_p}=\dfrac{1}{1}=1:1

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