Find the de Broglie wavelength λλlambda for an electron moving at a speed of 1.00×106m/s1.00×106m/s. (Note that this speed is low enough that the classical momentum formula p=mvp=mv is still valid.) Recall that the mass of an electron is me=9.11×10−31kgme=9.11×10−31kg, and Planck's constant is h=6.626×10−34J⋅sh=6.626×10−34J⋅s.

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explanation:

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speed of light in vacuum is 300,000 km/s

speed of light in other mediums is given by the expression v = c/n, where c is the speed of light in vacuum and n is the refractive index of medium.

for the first medium v = 200,000 km/s

  so, refractive index of medium = 300000/200000 = 1.5

    1.5 refractive index is usually for glass. so first medium is glass.

for the second medium v = 250,00 km/s

so, refractive index of medium = 300000/250000 = 1.2

    1.2 refractive index is usually for water. so second medium is water.

for the third medium v = 0 km/s

so, refractive index of medium = 300000/0 = ∞

    ∞ refractive index is usually for opaque mediums. so third medium is opaque medium.