Find the de broglie wavelength λ for an electron moving at a speed of 1.00×106m/s. (note that this speed is low enough that the classical momentum formula p=mv is still valid.) recall that the mass of an electron is me=9.11×10−31kg, and planck's constant is h=6.626×10−34j⋅s. g

De-broglie wavelength for an electron is given by,

λ=h/mv

here, h= plank's constant= 6.626×10³⁴

m= mass of the electron= 9.11×10⁻³¹

v= velocity of the electron= 1.00×10⁶

Substituting the values,

λ=

λ= 727.3 m

Hence, the de-broglie wavelength of the electron is 727.3 m.

the small 0.38-kg slider is known to move from position a to position b along the vertical-plane slot. determine (a) the work done on the body by its weight and (b) the work done on the body by the spring. the distance r = 0.70 m, the spring modulus k = 154 n/m, and the unstretched length of the spring is 0.53 m

explanation:

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