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SVETLANKA909090 [29]
3 years ago
13

An electron is confined to a one dimensional infinite potential well 150 pm. How much energy must it absorb if it is to jump to

the state n-3 from ground state?
Physics
1 answer:
igor_vitrenko [27]3 years ago
7 0

Answer:

\Delta E=1.22\times 10^{-22}J

Explanation:

The energy of electron in any state is given by E=\frac{n^2h^2}{8mL^2} here h is planck's constant n is state of electron L is the infinte potential well m is the mass of electron

We know that h=6.67\times 10^{-34}

Potential well dimension = 150pm=150\times 10^{-12}m

Mass of electron =9.1\times 10^{-31}kg

So energy required to electron to jump from ground state to 3rd state

\Delta E=\frac{h^2}{8mL^2}\left ( 3^2-1^2 \right )

\Delta E=\frac{\left ( 6.67\times 10^{-34} \right )^2}{8\times 9.1\times 10^{-31}(150\times 10^{-12})^2}\left ( 9-1 \right )

\Delta E=1.22\times 10^{-22}J

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g A wheel of diameter 8.0 cm has a cord of length 6.0 m wound around its periphery. Starting from rest, the wheel is given a con
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To solve this problem it is necessary to apply the equations related to the description of the tangential and angular movement.

The displacement where the speed and acceleration is related is given by the equation:

x = v_0t+\frac{1}{2}at^2

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v_0 = Initial velocity (0 because start from rest)

t = time

a = Acceleration

We have angular acceleration but not tangential acceleration. Tangential acceleration can be obtained through the relationship

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a = \frac{0.08}{2}(3)

a = 0.12m/s^2

And we have also that the displacement is

x = 6m

Now replacing,

x = v_0t+\frac{1}{2}at^2

6 = 0*t+\frac{1}{2}(0.12)t^2

6 = \frac{1}{2}(0.12)t^2

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7 0
3 years ago
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Estimate the speed of the water free surface and the time required to fill with water a cone-shaped container 1.5 m high and 1.5
Zolol [24]

Answer:

speed of water is 0.0007138m/s

Explanation:

From the law of conservation of mass

Rate of mass accumulation inside vessel = mass flow in - mass flow out

so, dm/dt = mass flow in - mass flow out

taking p as density

d \frac{dQ}{dt} = pq_i_n

where,

q(in) is the volume flow rate coming in

Q = is the volume of liquid inside tank at any time

But,

dQ = Adh

where ,

A = area of liquid surface at time t

h = height from bottom at time t

A = πr²

r is the radius of liquid surface

r = (1.5/2) \div 1.5 h = \frac{h}{2}

Hence,

\pi( \frac{h}{2} )^2\frac{dh}{dt} =q_i_n

\frac{dh}{dt} = \frac{q_i_n}{\pi (\frac{h}{2})^2 } =\frac{4q_i_n}{\pi h^2}

so, the speed of water surface at height h

v = \frac{dh}{dt} =\frac{4q_i_n}{\pi h^2}

where,

q_i_n is 75.7 L/min = 0.0757m³/min

h = 1.5m

so,

v = \frac{4 \times 0.0757}{\pi \times 1.5^2} \\\\v = 0.04283m/min

v = 0.04283 /60

v = 0.0007138m/s

Hence, speed of water is 0.0007138m/s

5 0
3 years ago
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