A particle of mass 9.1 times 10^{-31} , text{ | vedclass

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(D) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.For the particle:$\lambda_{pa} = \frac{h}{m v} = \frac{

Vedclass · Accepted Answer

$910$

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(D) The de-Broglie wavelength of a particle is given by $\lambda = \frac{h}{p} = \frac{h}{mv}$.For the particle:$\lambda_{pa} = \frac{h}{m v} = \frac{h}{9.1 	imes 10^{-31} 	imes 10^{6}} = \frac{h}{9.1 	imes 10^{-25}} \quad (i)$For the photon:$\lambda_{ph} = \frac{h}{p} = \frac{h}{10^{-27}} \quad (ii)$Dividing equation $(ii)$ by equation $(i)$:$\frac{\lambda_{ph}}{\lambda_{pa}} = \frac{h / 10^{-27}}{h / (9.1 	imes 10^{-25})} = \frac{9.1 	imes 10^{-25}}{10^{-27}}$$\frac{\lambda_{ph}}{\lambda_{pa}} = 9.1 	imes 10^{2} = 910$Thus,the wavelength of the photon is $910$ times the wavelength of the particle.

$A$ particle of mass $9.1 \times 10^{-31} \, \text{kg}$ travels in a medium with a speed of $10^{6} \, \text{m/s}$ and a photon of radiation with linear momentum $10^{-27} \, \text{kg} \cdot \text{m/s}$ travels in vacuum. The wavelength of the photon is $....$ times the wavelength of the particle.

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