Temperature is usually measured in degrees with a(n)

A proton is confined to a space 1 fm wide (about the size of an atomic nucleus). What's the minimum uncertainty in its velocity?

Daniel [21]

Answer:

Minimum uncertainty in velocity of a proton, $\Delta v\ge 3.15\times 10^7\ m/s$

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, $\Delta x=10^{-15}\ m$

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

$\Delta p.\Delta x\ge \dfrac{h}{4\pi}$

Since, p = mv

$\Delta (mv).\Delta x\ge \dfrac{h}{4\pi}$

$m \Delta v.\Delta x\ge \dfrac{h}{4\pi}$

$\Delta v\ge \dfrac{h}{4\pi m\Delta x}$

$\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}$

$\Delta v\ge 3.15\times 10^7\ m/s$

So, the minimum uncertainty in its velocity is greater than $3.15\times 10^7\ m/s$ . Hence, this is the required solution.

3 0

3 years ago

The Moon and Earth rotate about their common center of mass, which is located about RcM 4700 km from the center of Earth. (This

erica [24]

To solve this problem it is necessary to apply the concepts related to gravity as an expression of a celestial body, as well as the use of concepts such as centripetal acceleration, angular velocity and period.

PART A) The expression to find the acceleration of the earth due to the gravity of another celestial body as the Moon is given by the equation

$g = \frac{GM}{(d-R_{CM})^2}$