Question

The kinetic energy of a proton and that of an a-particle are 4 eV and 1 eV,
respectively. The ratio of the de-Broglie wavelengths associated with
them, will be
(a) 2:1
(b) 1:1
(c) 1:2
(d) 4:1
​

Answer 1

Answer:

• (b) 1:1

Explanation:

1. Formulae:

• 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

2. Reasoning

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)  

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