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 $\mathbf{1}\mathbf{.813}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{4}}$. By calculating its mass, identify the particle.

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 $1.813\times {10}^{-4}$.

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,

$m_e=9.109\times {10}^{-31}\text{\hspace{0.17em}kg}$

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

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

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

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

Then the equation for de Broglie for electron will be,

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

Let $\lambda$is de Broglie wavelength of an unknown particle.

Then,

$\lambda =\frac{h}{mv}$ ….. (2)

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

$\frac{\lambda }{{\lambda }_e}=\frac{1}{1.813\times {10}^{-4}}$

$\frac{{\lambda }_e}{\lambda }=1.813\times {10}^{-4}$ ….. (3)

Now, take a ratio as below.

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

Substitute $1.813\times {10}^{-4}$ for $\frac{{\lambda }_e}{\lambda }$, $9.109\times {10}^{-31}\text{\hspace{0.17em}kg}$for $m_e$, and $\frac{1}{3}$for $\frac{v_e}{v}$in the above equation.

$\begin{array}{rcl}m& =& \frac{9.109\times {10}^{-31}\text{\hspace{0.17em}kg}}{3(1.813\times {10}^{-4})}\\ & =& 1.675\times {10}^{-27}\text{\hspace{0.17em}kg}\end{array}$

The mass of the unknown particle is $1.675\times {10}^{-27}\text{\hspace{0.17em}kg}$.

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

Hence, the particle is the neutron.