Question

The ratio between the de-Broglie wavelength associated with protons, accelerated through a potential of $512V$ and $\alpha$-particles accelerated through a potential of ${X_0}Volt$ is found to be $1$. Find the value of ${X_0}$.

Answer 1

We know that radiation has dual nature that is, both wave and particle nature. De-Broglie proposed that this is true for matter as well. He also said that frequency and wavelength are associated with an electron’s energy and momentum. We will be using the fact that charge of an $\alpha$-particle is two times the charge of a proton and mass of an $\alpha$-particle is four times the mass of a proton.

Formula used:
The formula,$\lambda = \dfrac{h}{{\sqrt {2meV} }}$ is used where $\lambda$ is the wavelength, $h$ is planck's constant which is equal to $6.63 \times {10^{ - 34}}{m^2}kg{s^{ - 1}}$ , $e$ is the charge of the particle and $V$ is the potential .

Complete step by step answer:
By using the formula $\lambda = \dfrac{h}{{\sqrt {2meV} }}$ for proton, we get
${\lambda _p} = \dfrac{h}{{\sqrt {2{m_p}e \times 512} }}$ where ${\lambda _p}$ is the de-Broglie wavelength of proton.
By using the formula $\lambda = \dfrac{h}{{\sqrt {2meV} }}$ for $\alpha$-particle, we get
${\lambda _\alpha } = \dfrac{h}{{\sqrt {2{m_\alpha }(2e) \times {X_0}} }}$ ($\because$ charge of an $\alpha$-particle is two times the charge of a proton) where ${\lambda _\alpha }$ is the de-Broglie wavelength of $\alpha$-particle.
It is given that the ratio between ${\lambda _p}$ and ${\lambda _\alpha }$ is equal to one.
$\dfrac{{{\lambda _p}}}{{{\lambda _\alpha }}} = 1$
$\Rightarrow \dfrac{{\sqrt {{m_\alpha } \times 2 \times {X_0}} }}{{\sqrt {{m_p} \times 512} }} = 1$
$\Rightarrow {X_0} = \sqrt {\dfrac{{{m_p} \times 256}}{{4({m_p})}}}$
$\therefore {X_0} = 8$

Hence,the value of ${X_0}$ is equal to $8$.

Additional information:
De Broglie waves account for the appearance of subatomic particles at conventionally unexpected sites because their waves penetrate barriers much as sound passes through walls. Thus a heavy atomic nucleus occasionally can eject a piece of itself in a process called alpha decay. The piece of nucleus (alpha particle) has insufficient energy as a particle to overcome the force barrier surrounding the nucleus; but as a wave it can leak through the barrier that is, it has a finite probability of being found outside the nucleus.

Note: Using the concept of the electron matter wave, de Broglie provided a rationale for the quantization of the electron’s angular momentum in the hydrogen atom, which was postulated in Bohr’s quantum theory. The physical explanation for the first Bohr quantization condition comes naturally when we assume that an electron in a hydrogen atom behaves not like a particle but like a wave.