If particles are moving with same velocity, then De-Broglie wavelength is maximum for :

Question

A photon and an electron have the same energy \( E \). If \( \lambda_p \) is the wavelength of the photon and \( \lambda_e \) is the de Broglie wavelength of the electron, then the ratio \( \frac{\lambda_p}{\lambda_e} \) is:

Answer 1

The question asks which particle has the maximum De-Broglie wavelength if particles are moving with the same velocity. To answer this, we need to use the concept of De-Broglie wavelength.

The De-Broglie wavelength \(\lambda\) of a particle is given by the formula:

\(\lambda = \frac{h}{mv}\)

where:

\(h\) is the Planck's constant.
\(m\) is the mass of the particle.
\(v\) is the velocity of the particle.

Since all particles are moving with the same velocity \(v\), the De-Broglie wavelength depends inversely on the mass of the particle.

Therefore, the De-Broglie wavelength will be maximum for the particle with the minimum mass.

Let's compare the masses of the given particles:

Proton: Mass approximately \(\approx 1 \, \text{amu}\)
Neutron: Mass approximately \(\approx 1 \, \text{amu}\)
\(\alpha\)-particle: Mass approximately \(\approx 4 \, \text{amu}\)
\(\beta\)-particle: This usually refers to an electron or positron, with the electron having a mass approximately \(\approx 0.00055 \, \text{amu}\)

Among these, the mass of a \(\beta\)-particle (electron) is the smallest.

Hence, the De-Broglie wavelength is maximum for the \(\beta\)-particle.

Answer 2