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Alenkinab [10]
3 years ago
6

. As we increase the quantum number of an electron in a one-dimensional, infinite potential well, what happens to the number of

maximum points in the probability density function?
It increases.

It decreases.

It remains the same
Physics
1 answer:
Natalija [7]3 years ago
8 0

Answer:

It increases.

Explanation:

For the electron to escape the photon needs energy is equal to the difference between initial and its non quantised region energy , then only it will be able to escape finite well.

E ∝ n^2

n= energy state quantum number

so if , n increases maximum point of probability density increases.

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In a double-slit interference experiment you are asked to use laser light of different wavelengths and determine the separation
MAXImum [283]

Answer:

Red light

Explanation:

This because All interference or diffraction patterns depend upon the wavelength of the light (or whatever wave) involved. Red light has the longest wavelength (about 700 nm)

3 0
3 years ago
4.
xxTIMURxx [149]
Idk what to say to this
3 0
3 years ago
you are given an orange liquid. what methos would you use to observe and describe the physcial properties of the liquid without
DedPeter [7]
Color, viscosity(thickness), smell, weight
3 0
3 years ago
Read 2 more answers
What fraction of all the electrons in a 25 mg water
mihalych1998 [28]

Answer:

9.11\times 10^{-15}.

Explanation:

The water droplet is initially neutral, it will obtain a 40 nC of charge when a charge of  -40 nC is removed from the water droplet.

The charge on one electron, \rm e=-1.6\times 10^{-19}\ C.

Let the N number of electrons have charge -40 nC, such that,

\rm Ne=-40\ nC\\\Rightarrow N=\dfrac{-40\ nC}{e}=\dfrac{-40\times 10^{-9}\ C}{-1.6\times 10^{-19}\ C}=2.5\times 10^{11}.  

Now, mass of one electron = \rm 9.11\times 10^{-31}\ kg.

Therefore, mass of N electrons = \rm N\times 9.11\times 10^{-31}=2.5\times 10^{11}\times 9.11\times 10^{-31}=2.2775\times 10^{-19}\ kg.

It is the mass of the of the water droplet that must be removed in order to obtain a charge of 40 nC.

Let it is m times the total mass of the droplet which is 25\ \rm mg = 25\times 10^{-6}\ kg.

Then,

\rm m\times (25\times 10^{-3}\ kg) = 2.2775\times 10^{-19}\ kg.\\m=\dfrac{2.2775\times 10^{-19}\ kg}{25\times 10^{-3}\ kg}=9.11\times 10^{-15}.

It is the required fraction of mass of the droplet.

3 0
3 years ago
1.Suppose someone pulls a cart up a ramp a distance of 85cm along the ramp with a force of 15N.
Drupady [299]

1. 12.75 J

Assuming that the force applied is parallel to the ramp, so it is parallel to the displacement of the cart, the work done by the force is

W=Fd

where

F = 15 N is the magnitude of the force

d = 85 cm = 0.85 m is the displacement of the cart

Substituting in the formula, we get

W=(15 N)(0.85 m)=12.75 J


2. 10.6 N

In this part, the cart reaches the same vertical height as in part A. This means that the same work has been done (because the work done is equal to the gain in gravitational potential energy of the object: but if the vertical height reached is the same, then the gain in gravitational potential energy is the same, so the work done must be the same).

Therefore, the work done is

W=Fd=12.75 J

However, in this case the displacement is

d = 120 cm = 1.20 m

Therefore, the magnitude of the force in this case is

F=\frac{W}{d}=\frac{12.75 J}{1.20 m}=10.6 N

3 0
3 years ago
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