The de Broglie wavelength of a proton is twic | vedclass

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(A) The de-Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and

Vedclass · Accepted Answer

$1: 1$

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(A) The de-Broglie wavelength $\lambda$ of a particle is given by $\lambda = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass,and $K$ is the kinetic energy.From this,we have $\lambda \propto \frac{1}{\sqrt{mK}}$,which implies $K \propto \frac{1}{m\lambda^2}$.Given that $\lambda_p = 2\lambda_\alpha$ and knowing the mass of an alpha particle $m_\alpha \approx 4m_p$:$\frac{K_p}{K_\alpha} = \frac{m_\alpha}{m_p} 	imes \left( \frac{\lambda_\alpha}{\lambda_p} ight)^2$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$$\frac{K_p}{K_\alpha} = 4 	imes \frac{1}{4} = 1$Therefore,the ratio of the kinetic energies 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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