Question

A proton, a neutron, an electron, and an $\alpha $-particle have the same energy. Then their de-Broglie wavelengths compare as:
$(i){\text{ }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ }} \succ {\text{ }}{\lambda _e}{\text{ }} \succ {\text{ }}{\lambda _\alpha }{\text{ }}$
$(ii){\text{ }}{\lambda _\alpha }{\text{ }} \prec {\text{ }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ }} \prec {\text{ }}{\lambda _e}{\text{ }}$
$(iii){\text{ }}{\lambda _e}{\text{ }} \prec {\text{ }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ }} \succ {\text{ }}{\lambda _\alpha }{\text{ }}$
$(iv){\text{ }}{\lambda _e}{\text{ = }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ = }}{\lambda _\alpha }{\text{ }}$

Answer 1

Hint: Using the formula of energy for the particle, we will derive a relation between the de Broglie wavelength and the mass of the given particles. Then by putting the mass of the given particle we get the de Broglie wavelength for each particle. Then we compare each wavelength by finding the ratio of each mass of each particle.
Formula Used:
$(i)$ Kinetic energy of particle $ = {\text{ }}\dfrac{1}{2}m{v^2}$
$(ii){\text{ }}\lambda {\text{ = }}\dfrac{h}{{mv}}$

Complete Answer:
We know that the kinetic energy of moving particle is given by the relation,
Kinetic energy of particle $ = {\text{ }}\dfrac{1}{2}m{v^2}$
$K{\text{ = }}\dfrac{1}{2}m{v^2}$
Multiply both sides by m and taking square root of both sides we get,
$\sqrt {2Km} {\text{ = }}mv$ ____________$(1)$
Also, from the definition of de Broglie we can calculate the de Broglie wavelength as,
$\lambda {\text{ = }}\dfrac{h}{{mv}}$
Now using the value of mv from equation $(1)$ and putting here we get,
$\lambda {\text{ = }}\dfrac{h}{{\sqrt {2Km} }}$
Now here we can observe that the de Broglie wavelength is dependent on the mass and kinetic energy of the particle as,
$\lambda {\text{ }} \propto {\text{ }}\dfrac{1}{{\sqrt {2Km} }}$
Thus we can compare the different wavelengths of the given particle by putting their mass in the above equation. Also the value of kinetic energy is equal for all particles then we got final result as,
$\lambda {\text{ }} \propto {\text{ }}\dfrac{1}{{\sqrt m }}$
Since we know that, ${m_p}{\text{ = }}{{\text{m}}_n}$. Thus we can say that ${\lambda _p}{\text{ = }}{\lambda _n}$. Also ${m_\alpha }{\text{ }} \succ {\text{ }}{{\text{m}}_p}$, we can say that ${\lambda _\alpha }{\text{ }} \prec {\text{ }}{\lambda _p}$. Also ${m_e}{\text{ }} \prec {\text{ }}{{\text{m}}_n}$, hence we can say that ${\lambda _e}{\text{ }} \succ {\text{ }}{\lambda _n}$. Thus on comparing all the masses with their wavelength we get the order their de Broglie wavelength as,
${\lambda _\alpha }{\text{ }} \prec {\text{ }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ }} \prec {\text{ }}{\lambda _e}{\text{ }}$

Hence the correct option is $(ii){\text{ }}{\lambda _\alpha }{\text{ }} \prec {\text{ }}{\lambda _p}{\text{ = }}{\lambda _n}{\text{ }} \prec {\text{ }}{\lambda _e}{\text{ }}$.

Note:
When we find the relation between the wavelength and the mass of the particle we remove other constant quantities like Planck’s constant. The relation between de Broglie wavelength and mass is inverse. Therefore the particle which has higher mass will have a smaller de Broglie wavelength.