Question

calculate the de-Broglie associated with an electron with velocity 1.6 × 10^6 ms^-1?​

Answer 1

Answer:

Formulae :

λ=hmv

=6.625×10−349.1×10−31×1.6×106

=0.455×10−9m

=0.455 m.

v=1.6×106m/sec

h=6.625×10−34J.sec

m=9.1×10−31kg.

Explanation:

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Answer 2

The de Broglie wavelength associated with an electron with velocity $1.6 * 10^{6} ms^{-1}$ is $4.545 * 10^{-10}\ m.$

Explanation:

The formula for de Broglie wavelength is;

λ   = h/p

λ  = de Broglie wavelength

h = plank's constant

p = momentum.

Momentum can be defined as  $p = mv$

m = mass of the particle, an electron in this case.

v = velocity of particle.

After substituting the value of p the formula can be written as:

λ = h/mv

we know, the mass of electron = $9.11 * 10^{-31} kg$

h = $6.626 * 10^{-34} Js$

Given: v = $1.6 * 10^{6} ms^{-1}$

Substituting the values;

λ = $\frac{6.626 * 10^{-34} }{(9.11 * 10^{-31}) * (1.6 * 10^{6} )}$

λ = $4.545 * 10^{-10}\ m$

The  de Broglie wavelength = $4.545 * 10^{-10}\ m.$