Following particles have the same kinetic energy. Arrange them in the descending order of their de-Broglie wavelengths.
A. a proton
B. an \( \alpha \)-particle
C. a \( \beta \)-particle
D. a neutron
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → C, A, D, B
The de Broglie wavelength of a particle of kinetic energy \( K \) is
\( \displaystyle \lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}} \)
With the same \( K \) for all, \( \lambda \propto 1/\sqrt{m} \): the lighter the particle, the longer its wavelength.
C. a \( \beta \)-particle (electron): \( m \approx 9.1\times10^{-31}\,\mathrm{kg} \), the lightest, so the longest wavelength.
A. a proton: \( m_p = 1.6726\times10^{-27}\,\mathrm{kg} \)
D. a neutron: \( m_n = 1.6749\times10^{-27}\,\mathrm{kg} \), slightly heavier than the proton, so a slightly shorter wavelength.
B. an \( \alpha \)-particle: \( m \approx 4m_p \), the heaviest, so the shortest wavelength.
Descending order of wavelength: C, A, D, B.