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(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2Em}}$,where $h$ is Planck's constant,$E$ is the kinetic ene

Vedclass · Accepted Answer

$\lambda_{\alpha} < \lambda_{n} < \lambda_{p} < \lambda_{e}$

Vedclass · Answer

(B) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{\sqrt{2Em}}$,where $h$ is Planck's constant,$E$ is the kinetic energy,and $m$ is the mass of the particle.Since the energy $E$ is the same for all particles,we have $\lambda \propto \frac{1}{\sqrt{m}}$.The masses of the particles are related as $m_{e} Specifically,$m_{p} \approx m_{n}$ and $m_{\alpha} \approx 4m_{p}$.Since $\lambda$ is inversely proportional to the square root of mass,the particle with the smallest mass will have the largest wavelength.Therefore,$\lambda_{e} > \lambda_{p} \approx \lambda_{n} > \lambda_{\alpha}$.

$A$ proton,a neutron,an electron,and an $\alpha$-particle have the same energy. If $\lambda_{p}, \lambda_{n}, \lambda_{e},$ and $\lambda_{\alpha}$ are the de Broglie wavelengths of the proton,neutron,electron,and $\alpha$-particle respectively,then choose the correct relation from the following:

Similar Questions

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