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Q.

The de-Broglie wavelength of a particle moving with a velocity 2.25 × 10⁸ m/s is equal to the wavelength of photon. The ratio of kinetic energy of the particle to the energy of the photon is (velocity of light is 3 × 10⁸ m/s)

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a

$\frac{1}{8}$

b

$\frac{3}{8}$

c

$\frac{5}{8}$

d

$\frac{7}{8}$

answer is B.

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Detailed Solution

$K_{p a r t i c l e} = \frac{1}{2} m v^{2} A l s o λ = \frac{h}{m v} \Rightarrow K_{p a r t i c l e} = \frac{1}{2} (\frac{h}{λ v}) . v^{2} = \frac{v h}{2 λ} K_{p h o t o n} = \frac{h c}{λ} ∴ \frac{K_{p a r t i c l e}}{K_{p h o t o n}} = \frac{v}{2 c} = \frac{2.25 \times 10^{8}}{2 \times 3 \times 10^{8}} = \frac{3}{8}$

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