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Mademuasel
10 days ago
6

The energy gaps between the valence and conduction bands are called band gaps. For silicon, the band gap is 1.1 eV; for fused si

lica glass, it is 9.3 eV.
What is the wavelength λ of a photon that has energy 1.1eV?
The cut off between visible and infrared light is usually said to be somewhere between 700 and 800nm. Why is silicon transparent to most infrared light but opaque to visible light?
a) Visible photons have greater energy than the gap, so they can be absorbed whereas infrared photons pass through.
b) Visible photons have greater energy than the gap, so they can’t interact with the silicon as the infrared photons can.
c) Infrared photons have less energy than the gap, and so, unlike visible photons, they can be absorbed and reemitted from the material.
d) Infrared photons have less energy than the gap, and so they are only partially absorbed whereas visible photons are fully absorbed.
Physics
1 answer:
serg [3.4K]10 days ago
6 0
I appreciate you bringing your question here. I trust this response will be beneficial. Don’t hesitate to reach out with additional queries.
<span>d) Infrared photons have insufficient energy to bridge the gap, which leads to only partial absorption, in contrast to how visible photons are entirely absorbed.</span>
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Two of the types of ultraviolet light, uva and uvb, are both components of sunlight. their wavelengths range from 320 to 400 nm
Sav [3045]

In terms of light energy, a higher frequency corresponds to increased energy within the light.

We establish that frequency is essentially the inverse of wavelength:

frequency = 1 / wavelength

Calculating frequencies:

f UVA = 1/320 to 1/400

f UVA = 0.0031 to 0.0025

 

f UVB = 1/290 to 1/320

f UVB = 0.0034 to 0.0031

Since UVB occupies a higher frequency range, it consequently possesses greater energy than UVA.

7 0
22 days ago
Read 2 more answers
A solid metal sphere of diameter D is spinning in a gravity-free region of space with an angular velocity of ωi. The sphere is s
ValentinkaMS [3340]

Answer:

0.6

Explanation:

The formula for the volume of a sphere is \frac{4}{3} \pi (\frac{D}{2})^3

Thus \pi * r^2 * (\frac{D}{2} ) = \frac{4}{3} \pi (\frac{D}{2})^3

The radius of the disk is 1.15(\frac{ D}{2} )

Applying angular momentum conservation;

The M_i of the sphere = \frac{2}{5} m \frac{D}{2}^2

M_i of the disk = m*\frac{ \frac{1.15*D}{2}^2 }{2}

\frac{wd}{ws} = \frac{\frac{2}{5}m * \frac{D}{2}^2}{ m * \frac{(\frac{`.`5*D}{2})^2 }{2} }

= 0.6

5 0
1 month ago
Rohit placed a pencil perpendicular to principal axis in front of a converging
serg [3462]

Response:

45cm

Clarification:

A converging mirror is generally termed a concave mirror. The focal length and the image distance for a concave mirror are both expressed as positive values.

Using the mirror formula to derive the object distance;

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

Where f denotes the focal length, u indicates the object distance, and v represents the image distance.

Given f = 30cm, and v = 2u (The formed image is double the size of the pencil)

Plugging these values into the formula to solve for u yields;

\frac{1}{30} = \frac{1}{u} + \frac{1}{2u}\\\frac{1}{30} = \frac{2+1}{2u}\\\frac{1}{30} = \frac{3}{2u}\\

By cross-multiplying, we obtain;

2u = 90

Dividing both sides by 2;

2u/2 = 90/2

u = 45cm

The object's distance from the mirror measures 45cm

3 0
1 month ago
5. The speed of a transverse wave on a string is 170 m/s when the string tension is 120 ????. To what value must the tension be
Sav [3045]

Answer:

The tension in the string when the speed increased is 134.53 N

Explanation:

Given;

Tension in the string, T = 120 N

initial speed of the transverse wave, v₁ = 170 m/s

final speed of the transverse wave, v₂ = 180 m/s

The wave speed is expressed as;

v = \sqrt{\frac{T}{\mu} }

where;

μ represents mass per unit length

v^2 = \frac{T}{\mu} \\\\\mu = \frac{T}{v^2} \\\\\frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}

The new tension T₂ will be computed as;

T_2 = \frac{T_1 v_2^2}{v_1^2} \\\\T_2 = \frac{120*180^2}{170^2} \\\\T_2 = 134.53 \ N

Consequently, the tension in the string when the speed was increased is 134.53 N

3 0
20 days ago
A Chevrolet Corvette convertible can brake to a stop from a speed of 60.0 mi/h in a distance of 123 ft on a level roadway. What
kicyunya [3154]

Response:

83.1946504051 m

Rationale:

u = Starting velocity = 60\ mph=\dfrac{60\times 1609.34}{3600}=26.82233\ m/s

s = Distance traveled = 123\ ft=\dfrac{123}{3.281}=37.4885705578\ m

\theta = Incline = 26^{\circ}

v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-26.82233^2}{2\times 37.4885705578}\\\Rightarrow a=-9.5954230306\ m/s^2

Friction coefficient

\mu=-\dfrac{a}{g}\\\Rightarrow \mu=\dfrac{9.5954230306}{9.81}\\\Rightarrow \mu=0.978126710561

mg sin\theta - u mg cos\theta = ma\\\Rightarrow a=9.81(sin26-0.978126710561cos26)\\\Rightarrow a=-4.32382\ m/s^2

v^2-u^2=2as\\\Rightarrow s=\dfrac{v^2-u^2}{2a}\\\Rightarrow s=\dfrac{0^2-26.82233^2}{2\times -4.32382}\\\Rightarrow s=83.1946504051\ m

The calculated stopping distance is 83.1946504051 m

6 0
1 month ago
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