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(D) The kinetic energy of a particle is given by $K.E. = \frac{3}{2} KT = \frac{1}{2} mv^2$. Thus,$v = \sqrt{\frac{3KT}{m}}$. The de Broglie wavelengt

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Visible photon $>$ Thermal electron $>$ Thermal neutron

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(D) The kinetic energy of a particle is given by $K.E. = \frac{3}{2} KT = \frac{1}{2} mv^2$. Thus,$v = \sqrt{\frac{3KT}{m}}$. The de Broglie wavelength is $\lambda = \frac{h}{mv} = \frac{h}{m \sqrt{\frac{3KT}{m}}} = \frac{h}{\sqrt{3KTm}}$. This implies $\lambda \propto \frac{1}{\sqrt{m}}$. Since the mass of a photon is effectively zero (or it travels at $c$),it has the largest wavelength. Comparing particles,$m_{	ext{electron}} Therefore,$\lambda_{	ext{electron}} > \lambda_{	ext{neutron}} > \lambda_{	ext{proton}}$. Considering the photon,the order is $\lambda_{	ext{photon}} > \lambda_{	ext{electron}} > \lambda_{	ext{neutron}}$.

At temperature $T$,the average kinetic energy of any particle is $\frac{3}{2} KT$. The de Broglie wavelength follows the order:

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