ENGINEERING COLLEGE

What is the De Broglie wavelength of an electron under 150 V acceleration?

Answers

Answer 1

Answer:

Answer:

0.1 nm

Explanation:

Potential difference of the electron = 150 V

Mass of electron $m=9.1* 10^(-31)kg$

Charge on electron $1.6* 10^(-19)C$

Plank's constant $h=6.67* 10^(-34)$

If the velocity of the electron is v

Then according to energy conservation $eV =(1)/(2)mv^2$

$v=\sqrt{(2eV)/(m)}=\sqrt{(2* 1.6* 10^(-`19)* 150)/(9.1* 10^(-31))}=7.2627* 10^(6)m/sec$

According to De Broglie $\lambda =(h)/(mv)=(6.67* 10^(-34))/(9.1* 10^(-31)* 7.2627* 10^(6))=0.1nm$

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Calculate the "exact" alkalinity of the water in Problem 3-2 if the pH is 9.43.Calculate the "approximate" alkalinity (in mg/L as CaCO3 ) of a water containing 120 mg/L of bicarbonate ion and 15 mg/L of carbonate ion.

Answers

Answer:

A) approximate alkalinity = 123.361 mg/l

B) exact alkalinity = 124.708 mg/l

Explanation:

Given data :

A) determine approximate alkalinity first

Bicarbonate ion = 120 mg/l

carbonate ion = 15 mg/l

Approximate alkalinity = ( carbonate ion ) * 50/30 + ( bicarbonate ion ) * 50/61

= 15 * (50/30) + 120*( 50/61 ) = 123.361 mg/l as CaCO3

B) calculate the exact alkalinity of the water if the pH = 9.43

pH + pOH = 14

9.43 + pOH = 14. therefore pOH = 14 - 9.43 = 4.57

[OH^- ] = 10^-4.57 = 2.692*10^-5 moles/l

[ OH^- ] = 2.692*10^-5 * 179/mole * 10^3 mg/g = 0.458 mg/l

[ H^+ ] = 10^-9.43 * 1 * 10^3 = 3.7154 * 10^-7 mg/l

therefore the exact alkalinity can be calculated as

= ( approximate alkalinity ) + ( [ OH^- ] * 50/17 ) - ( [ H^+ ] * 50/1 )

= 123.361 + ( 0.458 * 50/17 ) - ( 3.7154 * 10^-7 * 50/1 )

= 124.708 mg/l

A tensile test is carried out on a bar of mild steel of diameter 20 mm. The bar yields under a load of 80 kN. It reaches a maximum load of 150 kN, and breaks finally at a load of 70 kN. Find (i) the tensile stress at the yield point (1i) the ultimate tensile stress; (iii) the average stress at the breaking point, if the diameter of the fractured neck is 10mm

Answers

Answer:

tensile stress at yield = 254 MPa

ultimate stress = 477 MPa

average stress = 892 MPa

Explanation:

Given data in question

bar yields load = 80 kN

load maximum = 150 kN

load fail = 70 kN

diameter of steel (D) = 20 mm i.e. = 0.020 m

diameter of breaking point (d) = 10 mm i.e. 0.010 m

to find out

tensile stress at the yield point , ultimate tensile stress and average stress at the breaking point

solution

in 1st part we calculate tensile stress at the yield point by this formula

tensile stress at yield = yield load / area

tensile stress at yield = 80 ×10³ / $\pi$ /4 × D²

tensile stress at yield = 80 ×10³ / $\pi$ /4 × 0.020²

tensile stress at yield = 254 MPa

in 2nd part we calculate ultimate stress by given formula

ultimate stress = maximum load / area

ultimate stress = 150 ×10³ / $\pi$ /4 × 0.020²

ultimate stress = 477 MPa

In 3rd part we calculate average stress at breaking point by given formula

average stress = load fail / area

average stress = 70 ×10³ / $\pi$ /4 × d²

average stress = 70 ×10³ / $\pi$ /4 × 0.010²

average stress = 892 MPa

A metal crystallizes with a face-centered cubic lattice. The edge of the unit cell is 408 pm. Calculate the number of atoms in the unit cell and diameter of the metal atom.

Answers

Answer:288 pm

Explanation:

Number of atoms(s) for face centered unit cell -

Lattice points: at corners and face centers of unit cell.

For face centered cubic (FCC), z=4.

- whereas

For an FCC lattices √2a =4r =2d

Therefore d = a/√2a = 408pm/√2a= 288pm

I think with this step by step procedure the, the answer was clearly stated.

A certain solar energy collector produces a maximum temperature of 100°C. The energy is used in a cyclic heat engine that operates in a 10°C environment. What is the maximum thermal efficiency? What is it if the collector is redesigned to focus the incoming light to produce a maximum temperature of 300°C?

Answers

Answer:

$\eta _(max) = 0.2413 = 24.13%$

$\eta' _(max) = 0.5061 = 50.61%$

Given:

$T_(1max) = 100^(\circ) = 273 + 100 = 373 K$

operating temperature of heat engine, $T_(2) = 10^(\circ) = 273 + 10 = 283 K$

$T_(3max) = 300^(\circ) = 273 + 300 = 573 K$

Solution:

For a reversible cycle, maximum efficiency, $\eta _(max)$ is given by:

$\eta _(max) = 1 - (T_(2))/(T_(1max))$

$\eta _(max) = 1 - (283)/(373) = 0.24$

$\eta _(max) = 0.2413 = 24.13%$

Now, on re designing collector, maximum temperature, $T_(3max)$ changes to $300^(\circ)$ , so, the new maximum efficiency, $\eta' _(max)$ is given by:

$\eta' _(max) = 1 - (T_(2))/(T_(3max))$

$\eta _(max) = 1 - (283)/(573) = 0.5061$

$\eta _(max) = 0.5061 = 50.61%$

Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,
b) sketch vc and ic.

Answers

Answer:

hello your question is incomplete attached is the complete question

A) Vc = 15 ( 1 - $e^(-t/0.15s)$ ) , ic = $1.5 mAe^(-t/0.15s)$

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

$R_(th)$ = 8 k ohms || 24 k ohms = 6 k ohms

$E_(th)$ = $(20 k ohms(20 v))/(24 k ohms + 8 k ohms)$ = 15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc = $E( 1 - e^(-t/t) )$

hence Vc = 15 ( 1 - $e^(-t/0.15)$ )

ic = $(E)/(R) e^(-t/t)$ = $(15)/(10) e^(-t/t)$ = $1.5 mAe^(-t/0.15s)$

attached

Shear modulus is analogous to what material property that is determined in tensile testing? (a)- Percent reduction of area (b) Yield strength (c)- Elastic modulus (d)- Poisson's ratio