The de-Broglie wavelength of a particle movin | vedclass

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(B) The kinetic energy of the particle is given by $K_{	ext{particle}} = \frac{1}{2}mv^2$. The de-Broglie wavelength is $\lambda = \frac{h}{mv}$,whic

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(B) The kinetic energy of the particle is given by $K_{	ext{particle}} = \frac{1}{2}mv^2$. The de-Broglie wavelength is $\lambda = \frac{h}{mv}$,which implies $mv = \frac{h}{\lambda}$.Substituting this into the kinetic energy equation: $K_{	ext{particle}} = \frac{1}{2} \left( \frac{h}{\lambda v} ight) v^2 = \frac{vh}{2\lambda}$  .......$(i)$The energy of a photon is given by $E_{	ext{photon}} = \frac{hc}{\lambda}$  .........$(ii)$Taking the ratio of the kinetic energy of the particle to the energy of the photon:$\frac{K_{	ext{particle}}}{E_{	ext{photon}}} = \frac{vh/2\lambda}{hc/\lambda} = \frac{v}{2c}$Substituting the given values $v = 2.25 	imes 10^8 \, m/s$ and $c = 3 	imes 10^8 \, m/s$:Ratio $= \frac{2.25 	imes 10^8}{2 	imes 3 	imes 10^8} = \frac{2.25}{6} = \frac{225}{600} = \frac{3}{8}$.

The de-Broglie wavelength of a particle moving with a velocity $2.25 \times 10^8 \, m/s$ is equal to the wavelength of a photon. The ratio of the kinetic energy of the particle to the energy of the photon is (velocity of light is $3 \times 10^8 \, m/s$):- (in $/8$)

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