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GrogVix [38]
3 years ago
13

Calculate the average linear momentum of a particle described by the following wavefunctions: (a) eikx, (b) cos kx, (c) e−ax2 ,

where in each one x ranges from −[infinity] to +[infinity].
Physics
1 answer:
Maksim231197 [3]3 years ago
6 0

Answer:

a) p=0, b) p=0, c) p= ∞

Explanation:

In quantum mechanics the moment operator is given by

              p = - i h’  d φ / dx

             h’= h / 2π

We apply this equation to the given wave functions

a)  φ = e^{ikx}

        .d φ dx = i k e^{ikx}

We replace

        p = h’ k e^{ikx}

        i i = -1

The exponential is a sine and cosine function, so its measured value is zero, so the average moment is zero

            p = 0

b) φ = cos kx

           p = h’ k sen kx

The average sine function is zero,

          p = 0

c) φ = e^{-ax^{2} }

         d φ / dx = -a 2x  e^{-ax^{2} }

         .p = i a g ’2x  e^{-ax^{2} }

       The average moment is

         p = (p₂ + p₁) / 2

         p = i a h ’(-∞ + ∞)

         p = ∞

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Answer:

Y-> + F-> +G ->=E->

Explanation:

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2 years ago
At locations A and B, the electric potential has the values VA = 1.83 V and VB = 5.17 V, respectively. A proton released from re
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Answer:

a. It starts at point B.

vp = 2.53*10⁴ m/s

a. it starts at point A.

ve= 1.08*10⁶ m/s

Explanation:

a)  As the proton is a positive charge, when released from rest, it will be accelerated due to the potential difference, from the higher potential to the lower one, so it is at the point B when released.

Once released, as the total energy must be conserved, the increase in kinetic energy must be equal (in magnitude) to the change in the electric potential energy, as follows:

ΔK + ΔUe = 0 ⇒ ΔK = -ΔUe =- (e*ΔV)

⇒ -( e* (VA-VB) ) = \frac{1}{2}*mp*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and mp= mass of proton = 1.67*10⁻²⁷ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{1.67e-27kg} } = 2.53e4 m/s

⇒ vp = 2.53*10⁴ m/s

b) If, instead of a proton, the charge realeased from rest, had been an electron, a few things would change:

First, as the electrons carry negative charges, they move from the lower potentials to the higher ones, which means that it would have started at point A.

Second, as its charge is (-e) the change in electric potential energy had been negative also:

ΔUe = -e*ΔV = -e* (VB-VA)

In order to find the speed of the electron when it is just passing point B, we can apply the conservation of energy principle as for the proton, as follows:

-( (-e)* (VB-VA) ) = \frac{1}{2}*me*v^{2}

where e= elementary charge= 1.6*10⁻¹⁹ C,  VA = 1.83 V, VB= 5.17V, and me= mass of electron = 9.1*10⁻³¹ kg.

Replacing by these values, and solving for v, we have:

v = \sqrt{\frac{2*1.6e-19C*3.34 V}{9.1e-31kg} } = 1.08e6 m/s

⇒ ve = 1.08*10⁶ m/s

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3 years ago
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Greyhound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but
maw [93]

{\large{\bold{\rm{\underline{Given \; that}}}}}

★ A grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

{\large{\bold{\rm{\underline{To\; find}}}}}

★ The speed of the hound and the hare

{\large{\bold{\rm{\underline{Solution}}}}}

★ The speed of the hound and the hare = 25:18

{\large{\bold{\rm{\underline{Full \; Solution}}}}}

\dashrightarrow  As it's given that a grey hound pursues a hare and takes 5 leaps for every 6 leaps of the hare, but 3 leaps of the hound are equal to 5 leaps of the hare.

 So firstly let us assume a metres as the distance covered by the hare in one leap.

Ok now let's talk about 5 leaps,.! As it's cleared that the hare cover the distance of 5a metres.

 But 3 leaps of the hound are equal to 5 leaps of the hare.

Henceforth, (5/3)a meters is the distance that is covered by the hound.

 Now according to the question,

Hound pursues a hare and takes 5 leaps for every 6 leaps of the hare..! (Same interval)

Now the distance travelled by the hound in it's 5 leaps..!

  • (5/3)a × 5

  • 25/3a metres

 Now the distance travelled by the hare in it's 6 leaps..!

  • 6a metres

 Now let us compare the speed of the hound and the hare. Let us calculate them in the form of ratio..!

  • 25/3a = 6a

  • 25/3 = 6

  • 25:18
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3 years ago
Light travels from medium X into medium Y. Medium Y has a higher index of refraction. Consider each statement below:(i) The ligh
ahrayia [7]

Answer:

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Explanation:

When light (electromagnetic radiation) travels through a material medium, its speed is less than the speed of light in a vacuum. If we define the index of parts

           n = c / v

where v is the speed of light in the material medium.

The direction of the ray can be determined by the law of refraction

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Let's apply these equations to our case where

          nₓ <n_y

i) The expression of the refractive index

          nₓ < n_y

         \frac{c}{v_x} = \frac{c}{n_y}

         v_y <vₓ

therefore the expression is FALSE

ii) If we use the law of refraction, the light, when passing from a medium with a lower start to another with a higher index, must approach the normal one, away from what would be the continuation of the path of the incident ray

so the expression is TRUE

iii) The speed of light is constant in all material media

the statement is FALSE

iv) light approaches normal

Let me clarify that the normal is a line perpendicular to the surface at the point of contact, not the direction of Io.

 the statement of FALSE

8 0
2 years ago
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