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kakasveta [241]
3 years ago
6

Neutrons travelling at 0.400 m/s are directed through a pair of slits having 1.00 m separation. an array of detectors is placed

10.0 m from the slits.
a) What is the de Broglie wavelength of the neutrons?
b) How far off axis if the first zero-intensity point on the detector array?
c) When a neutron reaches a detector, can we say which slit the neutrons passed through? Explain.
Physics
1 answer:
Tamiku [17]3 years ago
5 0

Answer:

9.91*10^-7 m

4.955*10^-6 m

Explanation:

Given that

v = 0.4 m/s

d = 1 m

L = 10 m

h = 6.62*10^-34 Js

m(neutron) = 1.67*10^-27 kg

To find the debroglie wavelength of the neutron, we use the formula

λ = h/mv

Now, we plug in the values we have listed.

λ = 6.62*10^-34 / (1.67*10^-27 * 0.4)

λ = 6.62*10^-34 / 6.68*10^-28

λ = 9.91*10^-7 m

b)

y1 = L (m + ½) λ/d, where m = 0

y1 = L (0 + ½) λ/d

y1 = L (½) λ/d

y1 = L/2 * λ/d or

y1 = Lλ/2d

now, we substitute the values for each of them, we have

y1 = (10 * 9.91*10^-7) / (2 * 1)

y1 = 9.91*10^-6 / 2

y1 = 4.955*10^-6 m

c) no, we can not say the neuron passed through one slit

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A car has a force of 2000N and a mass of<br> 1000kg. What is the acceleration of the<br> car?
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Answer:

100

Explanation:

by dividing 2000N and 1000kg.

5 0
4 years ago
A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon
Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

n=\frac{m}{M_m}

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

N=n N_A

where

n is the number of moles

N_A=6.022\cdot 10^{23} mol^{-1} is the Avogadro number

Substituting,

N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18} atoms

The energy emitted by each atom (the energy of one photon) is

E_1 = \frac{hc}{\lambda}

where

h is the Planck constant

c is the speed of light

\lambda=508 nm=5.08\cdot 10^{-7}nm is the wavelength

Substituting,

E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J

And so, the total energy emitted by the sample is

E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J

4 0
3 years ago
The Hall effect can be used to determine the density of mobile electrons in a conductor. A thin strip of the material being inve
solmaris [256]

Answer:

the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

Explanation:

Given the data in the question;

we make use of the following expression;

hall Voltage VH = IB / ned

where I = 2.25 A

B = 0.685 T

d =  0.107 mm =  0.107 × 10⁻³ m

e = 1.602×10⁻¹⁹ C

VH = 2.59 mV = 2.59 × 10⁻³ volt

n is the electron density

so from the form; VH = IB / ned

VHned = IB

n = IB / VHed

so we substitute

n = (2.25 × 0.685) / ( 2.59 × 10⁻³ × 1.602×10⁻¹⁹ × 0.107 × 10⁻³ )

n = 1.54125 /  4.4396226 × 10⁻²⁶

n = 3.4716 × 10²⁵ m⁻³

Therefore, the density of mobile electrons in the material is 3.4716 × 10²⁵ m⁻³

5 0
3 years ago
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