If all the particles have same kinetic energy, The relation between the wavelengths of alpha particle, electron and proton is:

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 determine the relation between the wavelengths of an alpha particle, an electron, and a proton, given that all have the same kinetic energy, we start by using the de Broglie wavelength formula:

\(\lambda = \frac{h}{\sqrt{2mE_k}}\)

Where:

\(h\) is the Planck's constant.
\(m\) is the mass of the particle.
\(E_k\) is the kinetic energy.

Since all three particles have the same kinetic energy, we can compare their wavelengths by looking at the mass \((m)\) of each particle:

The mass of an alpha particle (\(m_\alpha\)) is approximately 4 times the mass of a proton (comprising 2 protons and 2 neutrons).
The mass of a proton (\(m_p\)) is approximately 1,836 times the mass of an electron (\(m_e\)).

Using the de Broglie wavelength formula, we can establish the following order based on mass:

Electrons have the smallest mass, thus the highest wavelength.
Protons have an intermediate mass, thus an intermediate wavelength.
Alpha particles have the largest mass, thus the smallest wavelength.

Therefore, the correct order of wavelengths is:

\(\lambda_e > \lambda_p > \lambda_\alpha\)

Thus, the correct answer is: λ_e > λ_ρ > λ_α

If all the particles have same kinetic energy, The relation between the wavelengths of alpha particle, electron and proton is:

The Correct Option is B

Solution and Explanation

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