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

• (b) 1:1

Step-by-step 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:

`(m_\alpha)/(m_p)=4`

For the kinetic energies you find:

`(E_\alpha)/(E_p)=(m_\alpha * v_\alpha^2)/(m_p* v_p^2)`

`(1eV)/(4eV)=(4* v_\alpha^2)/(1* v_p^2)\\\\\\(v_p^2)/(v_\alpha^2)=16\\\\\\(v_p)/(v_\alpha)=4`

Thus:

`(m_\alpha)/(m_p)=4=(v_p)/(v__\alpha)`

`m_\alpha v_\alpha=m_pv_p`

From de-Broglie equation, λ = h/(mv)

`(\lambda_p)/(\lambda_\alpha)=(m_\lambda v_\lambda)/(m_pv_p)=(1)/(1)=1:1`