A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$ An alpha particle having certain kinetic energy has the same de-Brogle wavelength $\lambda$ The ratio of kinetic energy of proton and that of alpha particle is:
- $2: 1$
- $4: 1$
- $1: 4$
- $1: 2$
The Correct Option is B
Approach Solution - 1
The correct option is(B): 4:1
\(KE=\frac{p^2}{2m}=\frac{h^2}{2mλ^2} \)
\(\frac{KE_p}{KE_α}=\frac{m_α}{m_p}=4:1\)
Approach Solution -2
De-Broglie wavelength is given by: \[ \lambda = \frac{h}{p} \] Equating for both particles: \[ \frac{KE_p}{KE_\alpha} = \frac{m_\alpha}{m_p} = 4:1 \]
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Concepts Used:
Dual Nature of Radiation and Matter
The dual nature of matter and the dual nature of radiation were throughgoing concepts of physics. At the beginning of the 20th century, scientists untangled one of the best-kept secrets of nature – the wave-particle duplexity or the dual nature of matter and radiation.
Electronic Emission
The least energy that is needed to emit an electron from the surface of a metal can be supplied to the loose electrons.
Photoelectric Effect
The photoelectric effect is a phenomenon that involves electrons getting away from the surface of materials.
Heisenberg’s Uncertainty Principle
Heisenberg’s Uncertainty Principle states that both the momentum and position of a particle cannot be determined simultaneously.