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De Broglie postulated that for a particle with a momentum p, its corresponding wavelength is λ = n, where h is the Planck's constant. Assuming that a particle has a mass of m and a total kinetic energy of E, express the de Broglie wavelength of this particle as a function of E, m, and h only. (ii) Given that f(x) [In(5x2)]6, then determine the value of f'(2).

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jorgezarate

Answer:

(i) [tex]\lambda=\frac{h}{\sqrt{2mE_{k}}}[/tex]

(ii )[tex]f'(2)=1447.7[/tex]

Explanation:

(i)

de Broglie Equation:

λ = h/p  (1)

On the other hand, energy kinetic:

[tex]E_{k}=\frac{p^{2}}{2m}[/tex]

[tex]p=\sqrt{2mE_{k}}[/tex]  (2)

We replace (2) in  (1):

[tex]\lambda=\frac{h}{\sqrt{2mE_{k}}}[/tex]

(ii)

[tex]f(x)= (ln(5x^{2}))^{6}[/tex]

We derive the function:

[tex]f'(x)= 6(ln(5x^{2}))^{5}*\frac{1}{5x^{2}}*10x[/tex]

for x=2:

[tex]f'(2)=6(ln(5*2^{2}))^{5}*\frac{1}{5*2^{2}}*10*2=1447.7[/tex]

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