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Misha Larkins [42]

3 years ago

5

The bandgap of InP semiconductor laser is 1.0 eV. The effective mass of the valence band is ½ of the effective mass of the condu

ction band. Assuming that electron hole recombination transition occurs at 0.03 eV above the bandgap, calculate the wavelength of this transition.

1 answer:

klemol [59]3 years ago

4 0

Answer: the wavelength of this transition is 1.2039 um

Explanation:

Given that;

the energy level between the transitioning energy gap Eg = 1.0 + 0.03 = 1.03 eV

we know that λ = 1.24 / Eg

so we substitute our Eg into the above equation

λ = 1.24 / 1.03

λ = 1.2039 um

therefore the wavelength of this transition is 1.2039 um

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Very far from earth (at R- oo), a spacecraft has run out of fuel and its kinetic energy is zero. If only the gravitational force

Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

Rearrange this equation to find an equation for $v$ :

$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

It is already found that right before the collision, $\displaystyle \text{KE} = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ . Make use of this equation to find $v$ at that moment:

$\begin{aligned}v &= \sqrt{\frac{2\, \text{KE}}{m}} \\ &= \sqrt{\frac{2\, G\cdot M_\text{e} \cdot m}{R_\text{e}\cdot m}} = \sqrt{\frac{2\, G\cdot M_\text{e}}{R_\text{e}}}\end{aligned}$ .

6 0

3 years ago

What is happening when an oxygen molecule is formed from two separate oxygen atoms?

pickupchik [31]

Two from each oxygen atom. For examples of covalent bonding are water (H2O), and carbon dioxide (CO2). Covalent compounds are mostly gases or liquids with low melting points.

6 0

3 years ago

malfutka [58]

The answer for this is 1200N

6 0

3 years ago

The atmosphere is a mixture of gases. It is mostly composed of

katovenus [111]

Atmosphere has about 21% Oxygen , 78% Nitrogen and Some CO₂ and some other rare gases and water vapour.

So it is composed mostly of Nitrogen gas.

5 0

4 years ago

Where does an object need to be placed relative to a microscope for its 0.500 cm focal length objective to produce a magnificati

-BARSIC- [3]

Answer:

(a) $d_{o}= 0.501cm$ and (b) $d_{o}= 3.75cm$

Explanation:

The linear magnification (M) <u>produced by a lens is giving by</u>:

$M = - \frac{d_{i}}{d_{o}}$ (1)

<em>where $d_{i}$ : is the image distance and $d_{o}$ : is the object distance </em>

<u>Knowing that the</u> lens formula relates the focal length of an image with the distance of the image and the distance of the object:

$\frac{1}{f}= \frac{1}{d_{i}} + \frac{1}{d_{o}}$ (2)

(a) To calculate the distance of the object, first we have to express the equation (1) in function of the image distance:

$d_{i} = -M \cdot d_{o}$ (3)

Finally, by the introducing $d_{i}$ from (3) into equation (2) we can determine the distance of the object as follows:

$\frac{1}{f}= \frac{1}{-M \cdot d_{o}} + \frac{1}{d_{o}}$

$d_{o} = \frac {f (1 - M)}{-M}$ (4)

$d_{o} = \frac {0.500 cm (1 - (-400))}{-(-400)} = 0.501 cm$

So the object needs to be placed at 0.501 cm from the microscope.

(b) To find the distance of the length eyepiece <u>we will do the same treatment as before, to get to equation (4)</u>:

$d_{o} = \frac {f (1 - M)}{-M} = \frac {5.00 cm (1 - 4.00)}{-4.00} = 3.75 cm$

Have a nice day!

6 0

4 years ago