What should be the order of arrangement of de-Broglie wavelength of electron ($\lambda_e$), an $\alpha$-particle ($\lambda_\alpha$) and proton ($\lambda_p$) given that all have the same kinetic energy?

Question

Choose correct option :
(A) Number of photons required for a light beam of 2000 pm wavelength and 1 Joule energy is $1.01 \times 10^{16}$.
(B) Light with wavelength 300 nm has energy $E_1$ and for wavelength 900 nm has energy $E_2$, then $E_1/E_2 = 1/3$.
(C) Frequency of light is $4.5 \times 10^{16}$ Hz then its wavelength is $6.7 \times 10^{-9}$ m.
(D) If electrons and protons are accelerated by same potential difference, then their de-Broglie wavelength are equal.

Answer 1

To arrange the particles by their de-Broglie wavelength when they all have the same kinetic energy, we need to consider how the de-Broglie wavelength ($\lambda$) relates to the mass and velocity of each particle.

The de-Broglie wavelength is given by the formula:

$\lambda = \frac{h}{p}$, where $h$ is Planck's constant and $p$ is the momentum of the particle.

The momentum $p$ can be expressed as:

$p = \sqrt{2mK}$, where $m$ is the mass of the particle and $K$ is the kinetic energy.

Since all particles have the same kinetic energy, the momentum relation becomes:

$\lambda = \frac{h}{\sqrt{2mK}}$

Therefore, the de-Broglie wavelength $\lambda$ is inversely proportional to the square root of the mass of the particle:

$\lambda \propto \frac{1}{\sqrt{m}}$

Let's compare the masses:

Mass of an electron, $m_e$, is much smaller than that of a proton or an $\alpha$-particle.
Mass of a proton, $m_p$, is larger than $m_e$ but smaller than that of an $\alpha$-particle.
Mass of an $\alpha$-particle, $m_\alpha$, is approximately four times the mass of a proton as it consists of 2 protons and 2 neutrons.

This implies:

The de-Broglie wavelength of an electron ($\lambda_e$) will be the largest, since $\lambda \propto \frac{1}{\sqrt{m}}$, and it's the lightest particle.
The de-Broglie wavelength of a proton ($\lambda_p$) will be smaller than that of an electron, because it's heavier.
The de-Broglie wavelength of an $\alpha$-particle ($\lambda_\alpha$) will be the smallest as it's the heaviest.

Thus, the order of arrangement by decreasing de-Broglie wavelength will be:

$\lambda_e > \lambda_p > \lambda_\alpha$

This matches the correct answer.

What should be the order of arrangement of de-Broglie wavelength of electron ($\lambda_e$), an $\alpha$-particle ($\lambda_\alpha$) and proton ($\lambda_p$) given that all have the same kinetic energy?

Show Hint

The Correct Option is C

Solution and Explanation

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