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3 years ago

An electron has an uncertainty in its position of 587 pm . part a what is the uncertainty in its velocity?

1 answer:

6 0

We can solve the problem by using Heisenberg's uncertainty principle, which states that:
$\Delta x \Delta p \geq \frac{h}{4 \pi}$
where
$\Delta x$ is the uncertainty on the position
$\Delta p$ is the uncertainty on the momentum
$h$ is the Planck constant.

Keeping in mind that the momentum is the product between the mass of the electron and its velocity:
$p=mv$
we can rewrite the Heisenberg principle as
$m \Delta x \Delta v \geq \frac{h}{4 \pi}$
where $\Delta v$ is the uncertainty on the velocity.

The uncertainty on the position is $\Delta x = 587 pm = 587 \cdot 10^{-12} m$ , so we can find the uncertainty on the velocity by re-arranging the previous equation:
$\Delta v \geq \frac{h}{4 \pi m \Delta x}= \frac{6.6 \cdot 10^{-34} Js}{4 \pi (9.1 \cdot 10^{-31}kg)(587 \cdot 10^{-12} m)} =9.84 \cdot 10^4 m/s$

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If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you

Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, $Q=121\ nC=121\times 10^{-9}\ C$

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

$Q=ne$

e is the charge on electron

$n=\dfrac{Q}{e}$

$n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}$

$n=7.5625\times 10^{11}$

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0

3 years ago

Consider two sizes of disk, both of mass M. One size of disk has radius R; the other has radius 4R. System A consists of two of

Harman [31]

Answer:

4 smaller disks

Explanation:

We are given;

Mass of smaller and larger disks = M

Radius of smaller disk = R

Radius of larger disk = 4R

Formula for moment of inertia about cylinder axis is:

I = ½MR²

Thus;

For small disk, I_small = ½MR²

For large disk, I_large = ½M(2R)² = 2MR²

We are told that moment of inertia of System A consists of two of the larger disks. Thus;

I_A = 2 × I_large = 2 × 2MR²

I_A = 4MR²

We are also told that System B consists of one of the larger disks and a number of the smaller disks. Thus;

I_B = I_large + n(I_small)

Where n is the number of smaller disks.

I_B = 2MR² + n(½MR²)

I_B = MR²(2 + n/2)

We are told that the moment of inertia for system A equals the moment of inertia for system B. Thus;

I_A = I_B

So;

4MR² = MR²(2 + n/2)

MR² will cancel out to give;

4 = 2 + n/2

Multiply through by 2 to give;

8 = 4 + n

n = 8 - 4

n = 4

5 0

3 years ago

What are the benefits of solar cookers?

Elanso [62]

Answer:

Using clean, renewable, and readily available solar energy as fuel.

Preserving natural resources by not requiring the use of wood or other biomass fuels to cook.

Not producing dangerous emissions which pollute local environments and contribute to climate change.

Explanation:

5 0

2 years ago

Read 2 more answers

A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long

noname [10]

Answer:

Part a)

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

$M = 5.52 \times 10^{-5} H$

Part C)

$EMF = 0.1 V/s$

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

$B = \mu_0 n i$

n = number of turns per unit length

$n = \frac{N}{L}$

$n = \frac{350}{0.20}$

$n = 1750 turn/m$

now we know that magnetic field due to solenoid is

$B = (4\pi \times 10^{-7})(1750)(0.100)$

$B = 2.2 \times 10^{-4} T$

Now magnetic flux due to this magnetic field is given by

$\phi = B.A$

$\phi = (2.2 \times 10^{-4})(\pi r^2)$

$\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)$

$\phi = 2.76 \times 10^{-7} T m^2$

Part B)

Now for mutual inductance we know that

$\phi_{total} = M i$

$\phi_{total} = N\phi$

$\phi_{total} = 20(2.76 \times 10^{-4})$

$\phi_{total} = 5.52 \times 10^{-6}$

now we have

$M = \frac{5.52 \times 10^{-6}}{0.100}$

$M = 5.52 \times 10^{-5} H$

Part C)

As we know that induced EMF is given as

$EMF = M \frac{di}{dt}$

$EMF = 5.52 \times 10^{-5} (1800)$

$EMF = 0.1 V/s$

3 0

3 years ago

The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

german

Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of $CsCl$ :

(1) Conversion of solid calcium into gaseous cesium atoms.

$Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)$

$\Delta H_s$ = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

$Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)$

$\Delta H_I$ = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

$Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)$

$\Delta H_D$ = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

$Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)$

$\Delta H_L$ = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L$

$\Delta H_L=-667KJ/mole$

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

5 0

3 years ago