Question

The existence of the atomic nucleus was discovered in $\mathbf{\text{1911}}$by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold. (a) If the alpha particles had a kinetic energy of $\mathbf{7}\mathbf{.5}\mathbf{\text{MeV}}$, what wads their de Broglie wavelength? (b) Explain whether the wave nature of the incident alpha particles should have been taken into account in interpreting these experiments. The mass of an alpha particle is$\mathbf{\text{4.00u}}$ (atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about $\mathbf{\text{30fm}}$. (The wave nature of matter was not postulated until more than a decade after these crucial experiments were performed.)

Short answer

(a) The de Broglie wavelength is $5.2\text{fm}$.

(b) Wave nature of alpha particle need not to be taken into consideration.

Step-by-step solution

The given data:

The existence of the atomic nucleus was discovered in $\text{1911}$by Ernest Rutherford, who properly interpreted some experiments in which a beam of alpha particles was scattered from a metal foil of atoms such as gold.

The mass of an alpha particle is $4.00\text{u}$(atomic mass units), and its distance of closest approach to the nuclear center in these experiments was about $\text{30fm}$.

Mass of alpha particles, $m=4.00\text{u}$

The kinetic energy of alpha particles,$K=\text{7.5MeV}$

Concept and Formula used:

The de Broglie wavelength is given by

$\mathit{\lambda }\mathbf{=}\frac{\mathbf{h}\mathbf{c}}{\sqrt{\mathbf{2}\mathbf{m}{\mathbf{c}}^{\mathbf{2}}\mathbf{K}}}$

Here,$\mathbf{h}$ is Plank’s constant,$\mathbf{c}$is the speed of light, $\mathbf{m}$ is the mass of the alpha particle,$\mathbf{K}$is the kinetic energy of the particle.

(a) Find de Broglie wavelength:

Take the value of $hc$as,

$hc=1240\text{\hspace{0.17em}MeV}\cdot \text{fm}$

And the value of square of speed of light is,

$c^2=931.5\raisebox{1ex}{$\text{MeV}$}\!\left/ \!\raisebox{-1ex}{$\text{u}$}\right.$

The de Broglie wavelength is,

$\lambda =\frac{hc}{\sqrt{2m{c}^{2}K}}$

Substitute known values in the above equation.

$\begin{array}{rcl}\lambda & =& \frac{1240\text{\hspace{0.17em}MeV}\cdot \text{fm}}{\sqrt{2\left(4\text{\hspace{0.17em}u}\right)\left(\mathrm{9.31.5}\raisebox{1ex}{$\text{MeV}$}\!\left/ \!\raisebox{-1ex}{$\text{u}$}\right.\right)\left(7.5\text{\hspace{0.17em}MeV}\right)}}\\ & =& 5.2\text{\hspace{0.17em}fm}\end{array}$

Hence, de Broglie wavelength is $\begin{array}{rcl}& & 5.2\text{\hspace{0.17em}fm}\end{array}$.

(b) Find energy required by photon under photon microscope:

Since the wavelength,

$\begin{array}{rcl}\lambda & =& 5.2\text{\hspace{0.17em}fm}\\ & \ll & \text{30\hspace{0.17em}fm}\end{array}$

To a fair food approximation, the wave nature of the alpha particle does not need to be taken into consideration.