Respuesta :
Answer :
(a) The minimum uncertainty in an electron's velocity is, [tex]4.5\times 10^{4}m/s[/tex]
(b) The minimum uncertainty in a helium atom's velocity is, [tex]7.9\times 10^{1}m/s[/tex]
Explanation :
According to the Heisenberg's uncertainty principle,
[tex]\Delta x\times \Delta p=\frac{h}{4\pi}[/tex] ...........(1)
where,
[tex]\Delta x[/tex] = uncertainty in position
[tex]\Delta p[/tex] = uncertainty in momentum
h = Planck's constant
And as we know that the momentum is the product of mass and velocity of an object.
[tex]p=m\times v[/tex]
or,
[tex]\Delta p=m\times \Delta v[/tex] .......(2)
Equating 1 and 2, we get:
[tex]\Delta x\times m\times \Delta v=\frac{h}{4\pi}[/tex]
[tex]\Delta v=\frac{h}{4\pi \Delta x\times m}[/tex]
(a) Given:
m = mass of electron = [tex]9.11\times 10^{-31}kg[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\Delta x[/tex] = [tex]13\AA=13\times 10^{-10}m[/tex]
conversion used : [tex](1\AA=10^{-10}m)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (13\times 10^{-10}m)\times (9.1\times 10^{-31}kg)}[/tex]
[tex]\Delta v=4.5\times 10^{4}m/s[/tex]
The minimum uncertainty in an electron's velocity is, [tex]4.5\times 10^{4}m/s[/tex]
(b) Given:
m = mass of helium atom = [tex]6.646\times 10^{-27}kg[/tex]
h = Planck's constant = [tex]6.626\times 10^{-34}Js[/tex]
[tex]\Delta x[/tex] = [tex]1.0\AA=1.0\times 10^{-10}m[/tex]
conversion used : [tex](1\AA=10^{-10}m)[/tex]
Now put all the given values in the above formula, we get:
[tex]\Delta v=\frac{6.626\times 10^{-34}Js}{4\times 3.14\times (1.0\times 10^{-10}m)\times (6.646\times 10^{-27}kg)}[/tex]
[tex]\Delta v=7.9\times 10^{1}m/s[/tex]
The minimum uncertainty in a helium atom's velocity is, [tex]7.9\times 10^{1}m/s[/tex]
a) Minimum uncertainty in an electron's velocity is [tex]\rm 4.5\times 10^4 \;m/sec[/tex].
b) minimum uncertainty in a helium atom's velocity is 79 m/sec.
Solution :
a) Given :
[tex]\rm \Delta x = 13\; \AA = 13\times 10^-^1^0\;m[/tex]
Planck's Constant, [tex]\rm h = 6.626\times 10 ^-^3^4\; J.sec[/tex]
Mass of electron, [tex]\rm m_e = 9.11\times 10^-^3^1\;Kg[/tex]
Solution :
We know that mopmentum is,
[tex]\rm \Delta p = m \times \Delta v[/tex] --- (1)
Now according to Heisenberg's uncertainty principle,
[tex]\rm \Delta x \times \Delta p = \dfrac{h}{4\pi}[/tex] ---- (2)
So from equation (1) and (2) we get ,
[tex]\rm \Delta x \times \Delta v \times m = \dfrac{h}{2\pi}[/tex] ------ (3)
Now put the values of [tex]\rm m_e[/tex], h and [tex]\rm \Delta x[/tex] in equation (3) we get,
[tex]\rm \Delta v = \dfrac{6.626\times10^-^3^4}{4\pi\times 13\times 10^-^1^0\times 9.1 \times 10^-^3^1}[/tex]
[tex]\rm \Delta v = 4.5\times 10^4\;m/sec[/tex]
Therefore, minimum uncertainty in an electron's velocity is [tex]\rm 4.5\times 10^4 \;m/sec[/tex].
b) Given :
Mass of helium atom, [tex]\rm m_h_e = 6.646\times 10^-^2^7\; Kg[/tex]
[tex]\rm \Delta x = 1\AA= 10^-^1^0\;m[/tex]
Now again put the values of [tex]\rm m_h_e[/tex], h and [tex]\rm \Delta x[/tex] in equation (3) we get,
[tex]\rm \Delta v = \dfrac{6.626\times10^-^3^4}{4\pi\times 10^-^1^0\times 6.646 \times 10^-^2^7}[/tex]
[tex]\rm \Delta v = 79\;m/sec[/tex]
Therefore, minimum uncertainty in a helium atom's velocity is 79 m/sec.
For more information, refer the link given below
https://brainly.com/question/24508217?referrer=searchResults
