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4 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

1 answer:

Natalija [7]4 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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A car moves at a constant speed of 90km/h from a starting point. Another car moves at 70km/h after 2hours from the same starting

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Answer:

400

Explanation:

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3 years ago

Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.9 m tall window in 0.134 sec. From how high above th

timama [110]

Answer:

Explanation:

Given

length of window $h=2.9\ m$

time Frame for which rock can be seen is $\Delta t=0.134\ s$

Suppose h is height above which rock is dropped

Time taken to cover $h+2.9 is t_1$

so using equation of motion

$y=ut+\frac{1}{2}at^2$

where y=displacement

u=initial velocity

a=acceleration

t=time

time taken to travel h is

$h=0+0.5\times g\times (t_2)^2---2$

Subtract 1 and 2 we get

$2.9=0.5g(t_1^2-t_2^2)$

$5.8=g(t_1+t_2)(t_1-t_2))$

and from equation $t_1-t_2=0.134\ s$

so $t_1+t_2=\frac{5.8}{9.8\times 0.134}$

$t_1+t_2=4.416\ s$

and $t_1=t_2+\Delta t$

so $t_2+\Delta t+t_2=4.416$

$2t_2+0.134=4.416$

$t_2=0.5\times 4.282$

$t_2=2.141\ s$

substitute the value of $t_2$ in equation 2

$h=0.5\times 9.8\times (2.141)^2$

$h=22.46\ m$

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4 years ago

Two particles, each of mass m, are initially at rest very far apart.Obtain an expression for their relative speed of approach at

PSYCHO15rus [73]

Answer:

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

Explanation:

Consider two particles are initially at rest.

Therefore,

the kinetic energy of the particles is zero.

That initial K.E. = 0

The relative velocity with which both the particles are approaching each other is Δv and their reduced masses are

$\mu= \frac{m_1m_2}{m_1+m_2}$

now, since both the masses have mass m

therefore,

$\mu= \frac{m^2}{2m}$

= m/2

The final K.E. of the particles is

$KE_{final}=\frac{1}{2}\times \mu\times \Delta v^2$

Distance between two particles is d and the gravitational potential energy between them is given by

$PE_{Gravitational}= \frac{Gmm}{d}$

By law of conservation of energy we have

$KE_{initial}+KE_{final}= PE_{gravitaional}$

Now plugging the values we get

$0+\frac{1}{2}\frac{m}{2}\Delta v^2= -\frac{Gmm}{d}$

$|\Delta v |=\sqrt{\frac{4Gm}{d} }$

$=\sqrt{\frac{Gm}{d} }$

This the required relation between G,m and d

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3 years ago

I WILL GIVE BRAINLIEST TO WHOEVER IS CORRECT.

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The shorter the wavelength, the higher the frequency and energy.

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3 years ago

Read 2 more answers

Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge

NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

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for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as- $E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}$

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero between the two charges,but we can find the points situated on the axial line but outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining two charges where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

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3 years ago