91Ó°ÊÓ

A nonrelativistic particle is moving three times as fast as an electron.

The ratio of the de Broglie wavelength of the particle to that of the electron is $1.813×{10}^{−4}$.

Let $m_e$is mass of the electron and $v_e$is the speed of the electron, $m$is the mass of the unknown particle and role="math" localid="1663145451362" $v$is the speed of the unknown particle.

So,

$\frac{{\text{v}}_{\text{e}}}{\text{v}}=\frac{1}{3}$

And the mass of electron is,

$m_e=9.109×{10}^{−31}\text{ k²µ}$

The wavelength that is associated with an object in relation to its momentum and mass is known as the de Broglie wavelength. The de Broglie wavelength of a particle is usually inversely proportional to its strength.

The de Broglie wavelength is given by

$\begin{array}{rcl}\mathrm{λ}& =& \frac{h}{p}\\ & =& \frac{h}{mv}\end{array}$

Here, $h$is Plank’s constant, $p$ is the momentum, $m$is the mass of particle, $v$is the velocity of the particle.

Let ${\mathrm{λ}}_e$ is de Broglie wavelength of the electron.

Then the equation for de Broglie for electron will be,

${\mathrm{λ}}_e=\frac{h}{m_e{v}_e}$ ….. (1)

Let $\mathrm{λ}$is de Broglie wavelength of an unknown particle.

Then,

$\mathrm{λ}=\frac{h}{mv}$ ….. (2)

As given that, the ratio of the de Broglie wavelength of the particle to that of the electron is,

$\frac{\mathrm{λ}}{{\mathrm{λ}}_e}=\frac{1}{1.813×{10}^{−4}}$

$\frac{{\mathrm{λ}}_e}{\mathrm{λ}}=1.813×{10}^{−4}$ ….. (3)

Now, take a ratio as below.

$$\begin{gathered} \frac{\mathrm{λ}}{{\mathrm{λ}}_e}=\frac{m_e{v}_e}{mv} \\ m=\frac{v_e{\mathrm{λ}}_e}{v\mathrm{λ}}{m}_e \end{gathered}$$

Substitute $1.813×{10}^{−4}$ for $\frac{{\mathrm{λ}}_e}{\mathrm{λ}}$, $9.109×{10}^{−31}\text{ k²µ}$for $m_e$, and $\frac{1}{3}$for $\frac{v_e}{v}$in the above equation.

$\begin{array}{rcl}m& =& \frac{9.109×{10}^{−31}\text{ k²µ}}{3(1.813×{10}^{−4})}\\ & =& 1.675×{10}^{−27\text{ }}\text{ k²µ}\end{array}$

The mass of the unknown particle is $1.675×{10}^{−27}\text{ k²µ}$.

After checking in Appendix, it is clear that this is the mass of the neutron.

Hence, the particle is the neutron.