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2 years ago

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A beam of light is shined on a thin (sub-millimeter thick) single crystal wafer of material. The light source is special since i

t can be tuned to provide any wavelength of visible light on demand. The specimen is illuminated such that the wavelength of light is increased over time while the transmitted intensity of the light is measured. If the sample becomes transparent when the wavelength is greater than 690 nanometers, what is the band gap of the material, in eV

1 answer:

Soloha48 [4]2 years ago

7 0

Answer:

The energy gap of the material is $E_G = 1.7982 eV$

Explanation:

From the question we are told that

The wavelength is $\lambda = 690 nm$

The band gap of the material can be mathematically represented as

$E_G = \frac{h c}{\lambda }$

Where h is the Planck constant with value $h = 6.626 *10^{-34} joule \cdot sec$

c is the speed of light with a value $c = 3.0*10^8$

Substituting value

$E_G =\frac{6.626 *10^{-34} 3 *10^{8}}{690 *10^{-9}}$

$= 2.8809 *10^{-19}J$

Now converting to eV we divide by the charge on on electron. the value is

$e = 1.602 *10^{-19 } C$

so

$E_G = \frac{2..8809 *10^{-19}}{1.602 *10^{-19}}$

$E_G = 1.7982 eV$

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Answer:

D:

Explanation:

Conduction because the heat energy is transferring directly from the separate metals convection involves "currents" like in a liquid or a gas. When you turn on the stove with a pot of water on the burner the water at the bottom of the pot gets heated first almost immediately this water rises and cooler water falls. Basically the water is "stirring" or shifting, exchanging the heat energy this is why water heats so evenly.

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2 years ago

A particular metal has a work function of 1.05 eV. A light is shined onto this metal with a corresponding wavelength of 324 nm.

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Answer:

Velocity of electron will be $v=0.986\times 10^6m/sec$

Explanation:

We have given work function of metal $\Phi =1.05eV=1.05\times 1.6\times 10^{-19}J=1.68\times 10^{-19}J$

Wavelength of the light $\lambda =324nm=324\times 10^{-9}m$

So energy is given by $E=\frac{hc}{\lambda }$ , here h is plank's constant and c is speed of light

So $E=\frac{6.6\times 10^{-34}\times3\times 10^8}{324\times 10^{-9} }=6.11\times 10^{-19}j$

For a metal we know that $E=\Phi +KE_{MAX}$

So $KE_{MAX}=E-\Phi =6.11\times 10^{-19}-1.68\times 10^{-19}=4.43\times 10^{-19}$

Now kinetic energy is given by $KE=\frac{1}{2}mv^2$

$4.43\times 10^{-19}=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=0.986\times 10^6m/sec$

So velocity of electron will be $v=0.986\times 10^6m/sec$

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2 years ago

En el País Vasco los deportistas rurales levantan enormes piedras hasta su hombro. En un concurso , Jose levanta una piedra de 2

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Answer:

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

Explanation:

The sportsman that lifts the stone with a greater mass needs a higher force (El deportista que levanta la piedra con mayor masa necesita una mayor fuerza):

José

$F = (200\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 1961.4\,N$

Txomin

$F = (220\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)$

$F = 2157.54\,N$

Txomin lifted the stone with greater mass. (Txomin levantó la roca con mayor fuerza).

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2 years ago

A long, thin straight wire with linear charge density λ runs down the center of a thin, hollow metal cylinder of radius R. The c

netineya [11]

Answer:

$E=\frac{\lambda}{2\pi r\epsilon_0}$

Explanation:

We are given that

Linear charge density of wire= $\lambda$

Radius of hollow cylinder=R

Net linear charge density of cylinder= $2\lambda$

We have to find the expression for the magnitude of the electric field strength inside the cylinder r<R

By Gauss theorem

$\oint E.dS=\frac{q}{\epsilon_0}$

$q=\lambda L$

$E(2\pi rL)=\frac{L\lambda}{\epsilon_0}$

Where surface area of cylinder= $2\pi rL$

$E=\frac{\lambda}{2\pi r\epsilon_0}$

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2 years ago

Find the current that flows in a silicon bar of 10-μm length having a 5-μm × 4-μm cross-section and having free-electron and hol

klasskru [66]

The current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

<u>Explanation:</u>

Length of silicon bar, l = 10 μm = 0.001 cm

Free electron density, Ne = 104 cm^3

Hole density, Nh = 1016 cm^3

μn = 1200 cm^2 / V s

μр = 500 cm^2 / V s

The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.

J = Je + Jh

J = n qE μn + p qE μp

where, n and p are electron and hole densities.

J = Eq (n μn + p μp)

we know that E = V / l

So, J = (V / l) q (n μn + p μp)

J = (1.6 $\times$ 10^-19) / 0.001 (104 $\times$ 1200 + 1016 $\times$ 500)

J = 1012480 $\times$ 10^-16 A / m^2.

or

J = 1.01 $\times$ 10^-9 A / m^2

Current, I = JA

A is the area of bar, A = 20 μm = 0.002 cm

I = 1.01 $\times$ 10^-9 $\times$ 0.002 = 2.02 $\times$ 10^-12

So, the current flowing in silicon bar is 2.02 $\times$ 10^-12 A.

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2 years ago