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4 years ago

A line spectrum is produced when an electron moves from one energy level

Physics

1 answer:

dolphi86 [110]4 years ago

4 0

Answer:

Explanation:

A line spectrum in form of emission line is usually produced when an electron moves from higher energy level to lower energy orbit. During this process the electron loses certain amount of energy called photons with specific amount. The energy of the photon = the difference in energy of the orbits.

DE = E2-E1 where E2 electron with higher energy orbit.

It is possible for the electron to reach ground state in one jump or stop at one energy level temporarily but it cannot stop between energy levels.

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A 4-foot spring measures 8 feet long after a mass weighing 8 pounds is attached to it. The medium through which the mass moves o

brilliants [131]

Answer:

$\mathbf{x = 3te ^{-2 \sqrt{2t}}}$

Explanation:

From the given information concerning the spring-mass system:

Let us apply Hooke law.

Then, we have:

mg = ks

8 = k4

k = 8/4

k = 2

Provided that the mass weighing 8 lbs is attached to a spring.

Then, we can divide it by gravity 32 ft/s².

∴

m = 8/32

m = 1/4 slugs

The medium that offers the damping force $\beta = \sqrt{2}$

Now, let us set up a differential equation that explains the motion of the spring-mass system.

The general equation is:

$mx '' + \beta x' + kx = 0$

where;

$m = \dfrac{1}{4}$

k = 2, and

$\beta = \sqrt{2}$

Then;

$\dfrac{1}{4}x'' + \sqrt{2} x' + 2x = 0$

By solving the above equation, the auxiliary equation is:

$m^2 + 4 \sqrt{2} m + 8n = 0$

Using quadratic formula:

$\dfrac{-b \pm \sqrt{b^2 -4ac}}{2a}$

$m = \dfrac{-4 \sqrt{2} \pm \sqrt{(4 \sqrt{2})^2 -4(1)(8) }}{2}$

$m = \dfrac{-4 \sqrt{2} \pm \sqrt{32 -32 }}{2}$

$m = -2 \sqrt{2}$

Since this is a repeated root, the solution to their differential equation took the form.

$x_c = c_1 e^{mt} + c_2 t e^{mt}$

$x_c = c_1 e^{-2\sqrt{2} t} + c_2 t e^{-2\sqrt{2} t}$

From the initial condition.

At equilibrium position where the mass is being from:

x(0) = 0

Also, at the downward velocity of 3 ft/s

x'(0) = 3

Then, at the first initial condition:

$x_c (0) = c_1 e^{-2\sqrt{2} *0} + c_2 (0) e^{-2\sqrt{2} *0}$

$0= c_1 e^{0} + 0$

$0= c_1$

At the second initial condition;

$x' = -2 \sqrt{2} c_1 e^{-2 \sqrt{2} t } -2 \sqrt{2} c_2 t e^{-2 \sqrt{2} t } + c_2 e^{-2 \sqrt{2} t}$

where;

x'(0) = 3

$x' (0) = -2 \sqrt{2} c_1 e^{-2 \sqrt{2}* 0 } -2 \sqrt{2} c_2 (0) e^{-2 \sqrt{2} *0 } + c_2 e^{-2 \sqrt{2} *0}$

$3 = -2 \sqrt{2} * 0 *e^0 - 0 + c_2 e^0$

$3 = 0 + c_2$

$3 = c_2$

Replacing in the constraints, the equation of the motion is:

$\mathbf{x = 3te ^{-2 \sqrt{2t}}}$

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