The de Broglie wavelength of a proton is twic | vedclass

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Question

(A) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $m$ is the mass and $K$ is the kinetic energy

Vedclass · Accepted Answer

$1$:$1$

Vedclass · Answer

(A) The de Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $m$ is the mass and $K$ is the kinetic energy.From this,we have $\lambda \propto \frac{1}{\sqrt{mK}}$,which implies $K \propto \frac{1}{m\lambda^2}$.Let $m_p$ and $m_\alpha$ be the masses of the proton and alpha particle respectively. We know $m_\alpha = 4m_p$.Given $\lambda_p = 2\lambda_\alpha$.The ratio of kinetic energies is $\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p} 	imes \left( \frac{\lambda_\alpha}{\lambda_p} ight)^2$.Substituting the values: $\frac{K_p}{K_\alpha} = \left( \frac{4m_p}{m_p} ight) 	imes \left( \frac{\lambda_\alpha}{2\lambda_\alpha} ight)^2 = 4 	imes \left( \frac{1}{2} ight)^2 = 4 	imes \frac{1}{4} = 1$.Thus,the ratio is $1:1$.

The de Broglie wavelength of a proton is twice the de Broglie wavelength of an alpha particle. The ratio of the kinetic energies of the proton and the alpha particle is

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