Question

An electron, a doubly ionised Helium ion (\[H{e^{ + + }}\]) and a proton are having the same kinetic energy. The relation between their respective de-Broglie wavelengths \[{\lambda _e}\], \[{\lambda _{H{e^{ + + }}}}\], and \[{\lambda _p}\]is:
1. \[{\lambda _e} > {\lambda _p} > {\lambda _{H{e^{ + + }}}}\]
2. \[{\lambda _e} > {\lambda _{H{e^{ + + }}}} > {\lambda _p}\]
3. \[{\lambda _e} < {\lambda _p} < {\lambda _{H{e^{ + + }}}}\]
4. \[{\lambda _e} < {\lambda _{H{e^{ + + }}}} < {\lambda _p}\]

Answer 1

Hint: The electron is the lightest particle because it does not have any internal structure while the proton has an internal structure that consists of three quarks so it has a higher mass as compared to an electron.

Complete Step by Step Solution:
de-Broglie wavelength:
1. A wave is associated with every material object, whether it's a subatomic particle or a macro object, known as a matter-wave or de Broglie wave.
2. The de-Broglie wavelength equation is expressed as:
\[\lambda = \frac{h}{{m\upsilon }}\]
Where, \[\lambda \]is wavelength, \[h\]is plank’s constant (\[6.6262 \times {10^{ - 34}}Js\]), \[m\]is mass, and \[\upsilon \]is velocity

Relationship of de-Broglie wavelength based on same kinetic energy:
1. The relationship between the de-Broglie wavelength and the kinetic energy is expressed as:
\[\lambda = \frac{h}{{\sqrt {2m \times K.E.} }}\]
Where, K.E. is the kinetic energy
2. The mass of the doubled ionized Helium ion is \[6.6423 \times {10^{ - 27}}kg\], mass of a proton is \[1.673 \times {10^{ - 27}}kg\], and the mass of an electron is \[9.109 \times {10^{ - 31}}kg\].
3. The wavelength is inversely proportional to the mass so the wavelength is higher for less amount of mass which means the electron has a higher wavelength as compared to all proton and double ionized Helium ions.

Final answer:
Option (1) is correct.

Additional information:
1. The de-Broglie wavelength is presented on the basis of Einstein’s equation:
\[E = m{c^2}\]
2. According to Planck's equation every quantum of this wave has a discrete amount of energy, the wavelength of this wave is explained as:
\[E = h\upsilon \]
3. By comparison both the above equation,
\[m{c^2} = h\upsilon \]
4. The substitution of velocity in place of the speed of light because real particles do not travel at the speed of light
\[m{\upsilon ^2} = h\upsilon \]
5. The substitution of \[\frac{\upsilon }{\lambda }\]at \[\upsilon \]which relates the wavelength and particle with speed.
\[m{\upsilon ^2} = \frac{{h\upsilon }}{\lambda }\]

Note: Due to the fact that the momentum of macro objects is small compared to subatomic particles, the de Broglie wavelength of them is negligibly small, and thus we cannot see their wave nature.