Answer:
Part a)
[tex]\lambda = 0.345 m[/tex]
Part b)
[tex]\Delta x = 0.274 m[/tex]
Part c)
[tex]r = 2.8 \times 10^{11} m[/tex]
Explanation:
Part a)
De broglie wavelength is given as
[tex]\lambda = \frac{h}{mv}[/tex]
[tex]\lambda = \frac{1}{(0.145)(20)}[/tex]
[tex]\lambda = 0.345 m[/tex]
Part b)
By principle of uncertainty we know that
[tex]\Delta x \times \Delta P = \frac{h}{4\pi}[/tex]
[tex]\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}[/tex]
[tex]\Delta x = 0.274 m[/tex]
Part c)
As we know that
[tex]\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}[/tex]
also we know
[tex]mvr = \frac{nh}{2\pi}[/tex]
[tex]v = \frac{h}{2\pi mr}[/tex]
now we have
[tex]\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}[/tex]
[tex]r = \frac{h^2}{4\pi^2mke^2}[/tex]
[tex]r = 2.8 \times 10^{11} m[/tex]