Question

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).

Answer 1

(i)
`\lambda=\frac{h}{\sqrt{2mE_(k)}}`

(ii )
`f'(2)=1447.7`

Step-by-step explanation:

(i)

de Broglie Equation:

λ = h/p (1)

On the other hand, energy kinetic:

`E_(k)=(p^(2))/(2m)`

`p=\sqrt{2mE_(k)}` (2)

We replace (2) in (1):

`\lambda=\frac{h}{\sqrt{2mE_(k)}}`

(ii)

`f(x)= (ln(5x^(2)))^(6)`

We derive the function:

`f'(x)= 6(ln(5x^(2)))^(5)*(1)/(5x^(2))*10x`

for x=2:

`f'(2)=6(ln(5*2^(2)))^(5)*(1)/(5*2^(2))*10*2=1447.7`