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]
