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(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

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$\varepsilon_1 > \varepsilon_3 > \varepsilon_2$

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(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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