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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