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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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#86 can an object be moving when its acceleration is zero? can an object be accelerating when its speed is zero?

liraira [26]

Yes, an object<span> that was set in motion in the past by some force, but that is no longer being acted on by a net force, is </span>moving<span> but with </span>zero acceleration<span>, i.e. it is </span>moving<span> at constant velocity.</span>

7 0

3 years ago

Two long straight wires are parallel and 8.6 cm apart. They are to carry equal currents such that the magnetic field at a point

Neko [114]

Answer:

(a) The current should be in opposite direction

(b) The current needed is 39.8 A

Explanation:

Part (a)

Based, on right hand rule, the current should be in opposite direction

Part (b)

given;

strength of magnetic field, B = 370 µT

distance between the two parallel wires, d = 8.6 cm

$B = \frac{\mu_oI}{2\pi R}$

At the center, the magnetic field strength is twice

$B_c = 2(\frac{\mu_oI}{2\pi R}) =\frac{ \mu_oI}{\pi R}$

R = d/2 = 8.6/2 = 4.3 cm = 0.043 m

$B_c = \frac{ \mu_oI}{\pi R}\\\\I = \frac{B_c\pi R}{\mu_o} = \frac{370 *10^{-6}* \pi *0.043}{4\pi *10^{-7}}\\\\I = 39.8 \ A$

Therefore, current needed is 39.8 A

6 0

3 years ago

SOH-CAH-TOA is used to solve for the ________

Misha Larkins [42]

Answer:

c. initial (x and y)

Explanation:

When a projectile is launched at a velocity with a launch angle, to solve it, we must first resolve the initial velocity into the x and y components. To do this will mean we have to treat it like a triangle due to the launch angle and the direction of the projectile.

Therefore, we will have to make use of trigonometric ratios which is also known by the mnemonic "SOH CAH TOA"

Thus, this method resolves the initial x and y velocities.

3 0

3 years ago

A piston/cylinder contains 2 kg of water at 20◦C with a volume of 0.1 m3. By mistake someone locks the piston, preventing it fro

morpeh [17]

Answer:

Hi

Final temperature = 250.11 °C

Final volume = 0,1 m3.

Process work = 0

Explanation:

The specific volume in the initial state is: v = 0.1m3/2 kg = 0.05 m3/kg.

This volume is located between the volumes as saturated liquid and saturated steam at 20 °C. For this reason the water is initially in a liquid vapor mixture. As the piston was blocked the volume remains constant and the process is isometric, also known as isocoric process, so the final temperature will be the water temperature at a saturated steam of v=0.05m3/kg, which is obtained by using steam tables for water, by linear interpolation. As follows, using table A-4 of the Cengel book 7th Edition:

v=0.05 m3/kg

v1=0.057061 m3/kg

T1=242.56°C

v2=0.049779 m3/kg

T2=250.35°C

T= $\frac{T2-T1}{v2-v1} x(v-v1)+T1=\frac{250.35°C-242.56°C}{0.049779m3/kg-0.057061m3/kg}x(0.05m3/kg-0.057061m3/kg)+242.56°C=250.11°C$

The process work is zero because there is no change in volume during heating:

W=PxΔv=Px0=0

where

W=process work

P=pressure

Δv=change of volume, is zero because the piston was blocked so the volume remains constant.

7 0

3 years ago

In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin

alexandr402 [8]

Answer:

a) $v=0.999124c$

b) $E=7.566*10^{22}$

c) $E_a=760 times\ larger$

Explanation:

From the question we are told that

Distance to Betelgeuse $d_b=430ly$

Mass of Rocket $M_r=20000$

Total Time in years traveled $T_d=36years$

Total energy used by the United States in the year 2000 $E_{2000}=1.0*10^20$

Generally the equation of speed of rocket v mathematically given by

$v=\frac{2d}{\triangle t}$

$v=860ly/ \triangle t$

where

$\triangle t=\frac{\triangle t'}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\frac{36}{(\sqrt{1-860/ \triangle t)^2}}$

$\triangle t=\sqrt{(860)^2+(36)^2}$

$\triangle t=860.7532$

Therefore

$v=\frac{860ly}{ 860.7532}$

$v=0.999124c$

b)

Generally the equation of the energy E required to attain prior speed mathematically given by

$E=\frac{1}{\sqrt{1-(v/c)^2} }-1(20000kg)(3*10^8m/s)^2$

$E=7.566*10^{22}$

c)Generally the equation of the energy $E_a$ required to accelerate the rocket mathematically given by

$E_a=\frac{E}{E_{2000}}$

$E_a=\frac{7.566*10^{22}}{1.0*10^{20}}$

$E_a=760 times\ larger$

8 0

3 years ago