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2 years ago

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Given an electron beam whose electrons have kinetic energy of 10.0 kev , what is the minimum wavelength λmin of light radiated b

y such beam directed head-on into a lead wall? express your answer numerically in nanometers.

2 answers:

IgorLugansk [536]2 years ago

4 0

Answer: 0.1276 nm

Explanation:

$E=\frac{hc}{\lambda}$

E= energy of radiation $=10.0 kev= 1.6\times 10^{-15}J$

$1kev=1.6\times 10^{-16}Joules$

h = Planck's constant= $6.63\times 10^{-34}Js$

c = velocity of light $=3.08\times 10^{8}ms^{-1}$

$\lambda$ = wavelength of radiation = ?

$\lambda=\frac{hc}{E}$

$\lambda=\frac{6.63\times 10^{-34}Js\times 3.08\times 10^{8}ms^{-1}}{1.6\times 10^{-15}J}$

$\lambda=12.76\times 10^{-11}m=0.1270\times 10^{-9}m$

$\lambda=0.1276nm$

kolbaska11 [484]2 years ago

3 0

To answer the problem we would be using this formula which isE = hc/L where E is the energy, h is Planck's constant, c is the speed of light and L is the wavelength
L = hc/E = 4.136×10−15 eV·s (2.998x10^8 m/s)/10^4 eV

= 1.240x10^-10 m

= 1.240x10^-1 nm

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During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s

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Answer:

part a : <em>The dry unit weight is 0.0616 </em> $lb/in^3$ <em />

part b : <em>The void ratio is 0.77</em>

part c : <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}$

Here

$\gamma_d$ is the dry unit weight which is to be calculated
γ is the bulk unit weight given as

$\gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3$

w is the moisture content in percentage, given as 12%

Substituting values

$\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3$

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Part b

Void Ratio

The void ratio is given as

$e=\frac{G_s \gamma_w}{\gamma_d} -1$

Here

e is the void ratio which is to be calculated

$\gamma_d$ is the dry unit weight which is calculated in part a
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G is the specific gravity which is given as 2.72

Substituting values

$e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766$

<em>The void ratio is 0.77</em>

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Degree of Saturation

Degree of Saturation is given as

$S=\frac{G w}{e}$

Here

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G is the specific gravity which is given as 2.72

w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

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Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

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Here

$\gamma_{zav}$ is the zero air unit weight which is to be calculated

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$weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb$

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