Question

Kinetic energy of electron, proton and a particle is given as K, 2K and 4K respectively, then which of the following gives the correct order of De-Broglie wavelengths of electron, proton and a particle

• A. λp > λα > λe
• B. λα > λp > λe
• C. λe > λp > λα
• D. λe > λα > λp

Answer 1

The Correct Option is C

\(\lambda _{e}=\frac{h}{\sqrt{\frac{2\times m}{2000}}\times4K}=\frac{10\sqrt{20h}}{2\sqrt{2mK}}\)
\(\lambda _{p}=\frac{h}{\sqrt{2mK}}=\frac{h}{\sqrt{2mK}}\)
\(\lambda _{\alpha }=\frac{h}{\sqrt{2\times 4m\times 2K}}=\frac{h}{4\sqrt{mK}}\)
\(\lambda _{e}>\lambda _{p}>\lambda _{\alpha }\)

So, the correct option is (C): λe > λp > λα

Answer 2

The De Broglie wavelength of a particle is given by the equation:
λ = \(\frac{h}{p}\)
where λ is the wavelength, h is the Planck constant, and p is the momentum of the particle.
Since the kinetic energy of the particles is given, we can use the relation:
K = \(\frac{p^2}{2m}\)
where K is the kinetic energy, p is the momentum, and m is the mass of the particle.
Solving for momentum, we get:
p = \(\sqrt{(2mK)}\)
Using this expression for each particle and substituting in the equation for De Broglie wavelength, we get:
\(\lambda_e=\frac{h}{\sqrt{(2mK)}}\)
\(\lambda_p=\frac{h}{\sqrt{(4mK)}}\) = \(\frac{h}{\sqrt{(2m)}\sqrt{(2K)}}\)
\(\lambda_\alpha=\frac{h}{\sqrt{(4\alpha mK)}}\)
where α is the ratio of the mass of the alpha particle to the mass of the hydrogen atom.
Since the mass of the proton is approximately the same as the mass of the hydrogen atom, we can write:
α = \(\frac{m\alpha}{mH}=\frac{4}{1}\) = 4
Therefore, we have:
\(\lambda_e=\frac{h}{\sqrt{(2mK)}}\)
\(\lambda_p=\frac{h}{\sqrt{(2m)}\sqrt{(2k)}}\)
\(\lambda_\alpha=\frac{h}{\sqrt{(16mK)}}\)
Simplifying these expressions, we get:
\(\lambda_e=\frac{h}{\sqrt{(2mK)}}\)
\(\lambda_p=\frac{h}{\sqrt{(4mK)}}\)
\(\lambda_\alpha=\frac{h}{4\sqrt{mK}}\)
Since \(\sqrt{(2mK)}>\sqrt{(4mK)}>\frac{1}{(4\sqrt{mK})}\), 
We can conclude that: λe > λp > λα
Therefore, the correct option is (C).