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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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3. Maverick and Goose are flying a training mission in their F-14. They are
Elanso [62]

Answer:

A. The bomb will take <em>17.5 seconds </em>to hit the ground

B. The bomb will land <em>12040 meters </em>on the ground ahead from where they released it

Explanation:

Maverick and Goose are flying at an initial height of y_0=1500m, and their speed is v=688 m/s

When they release the bomb, it will initially have the same height and speed as the plane. Then it will describe a free fall horizontal movement

The equation for the height y with respect to ground in a horizontal movement (no friction) is

y=y_0 - \frac{gt^2}{2}    [1]

With g equal to the acceleration of gravity of our planet and t the time measured with respect to the moment the bomb was released

The height will be zero when the bomb lands on ground, so if we set y=0 we can find the flight time

The range (horizontal displacement) of the bomb x is

x = v.t     [2]

Since the bomb won't have any friction, its horizontal component of the speed won't change. We need to find t from the equation [1] and replace it in equation [2]:

Setting y=0 and isolating t we get

t=\sqrt{\frac{2y_0}{g}}

Since we have y_0=1500m

t=\sqrt{\frac{2(1500)}{9.8}}

t=17.5 sec

Replacing in [2]

x = 688\ m/sec \ (17.5sec)

x = 12040\ m

A. The bomb will take 17.5 seconds to hit the ground

B. The bomb will land 12040 meters on the ground ahead from where they released it

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How to solve arctan[tan(7pi /4)] ...?
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Well, 
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<span>arctan(tan(7pi/4)) = artan(tan(- pi/4)) = - pi/4 
</span>

I hope my answer has come to your help. Thank you for posting your question here in Brainly. We hope to answer more of your questions and inquiries soon. Have a nice day ahead!
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