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Question

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is t

Vedclass · Accepted Answer

$\varepsilon_1 > \varepsilon_3 > \varepsilon_2$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$,where $h$ is Planck's constant,$m$ is the mass of the particle,and $K$ is the kinetic energy.Since the de-Broglie wavelength $\lambda$ is the same for all three particles,we have $\sqrt{2mK} = 	ext{constant}$,which implies $mK = 	ext{constant}$ or $K \propto \frac{1}{m}$.The masses of the particles are $m_e$ (electron),$m_p$ (proton),and $m_{\alpha}$ (alpha particle). We know that $m_e Since $K$ is inversely proportional to mass,we have $\varepsilon_1 > \varepsilon_3 > \varepsilon_2$ (where $\varepsilon_1$ corresponds to the electron,$\varepsilon_3$ to the proton,and $\varepsilon_2$ to the alpha particle).

If the kinetic energies of an electron,an alpha particle,and a proton having the same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$,and $\varepsilon_3$ respectively,then:

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