Question

The de-Broglie wavelength ($\lambda$) for electron ($e$), proton ($p$), and He²⁺ ion ($\alpha$) are in the following order, Speed of $e$, $p$, and $\alpha$ are the same

• A. α>p>e
• B. e>p>α
• C. e>α>p
• D. αe

Answer 1

The Correct Option is B

The de-Broglie wavelength ($\lambda$) is given by the formula:

$\lambda = \dfrac{h}{mv}$ Where: - $h$ is Planck's constant - $m$ is the mass of the particle - $v$ is the velocity of the particle 

Since the speed of the electron ($e$), proton ($p$), and He²⁺ ion ($\alpha$) are the same, the de-Broglie wavelength is inversely proportional to the mass of the particle. 

- The electron ($e$) has the smallest mass, so it will have the largest de-Broglie wavelength. - The proton ($p$) has a larger mass than the electron, so its de-Broglie wavelength will be smaller. - The He²⁺ ion ($\alpha$) is the heaviest, so it will have the smallest de-Broglie wavelength. 

Thus, the order of de-Broglie wavelengths is: 

e > p > α 

Answer: e > p > α