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koban [17]
3 years ago
14

A scanning electron microscope (SEM) accelerates electrons to very high speeds and fires those electrons at a sample. By looking

at the way that the electrons interact with the sample, we can generate detailed images of objects we can't get with a traditional optical microscope, like mineral crystals or biological cells. To accomplish this goal, an SEM uses an accelerating voltage around 10kV (high voltage generally means better resolution). One such SEM uses an accelerating voltage of 13kV. How fast are the electrons moving after they pass through this accelerating voltage?
Physics
1 answer:
ss7ja [257]3 years ago
6 0

Answer:

v = 1.697 x 10^8 ms^-1

Explanation:

As we know ½ mv2 = eV where m is the mass of electron = 9 x 10^-31 kg and v is the velocity or speed of electrons….

We have V = 13kV = 1.3 x 10^-14 eV

By putting the values…

v2 = 2 eV / m = 2 x 1.3 x 10^-14 / 9 x 10^-31 = 2.88 x 10^ 16 ms^-1

v = 1.697 x 10^8 ms^-1

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Answer:

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Air at 80 kPa and 127 °C enters an adiabatic diffuser steadily at a rate of 6000 kg/h and leaves at 100 kPa. The velocity of the
Vinil7 [7]

Answer:

a) The exit temperature is 430 K

b) The inlet and exit areas are 0.0096 m² and 0.051 m²

Explanation:

a) Given:

T₁ = 127°C = 400 K

At 400 K, h₁ = 400.98 kJ/kg (ideal gas properties table)

The energy equation is:

q-w=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +delta-p

For a diffuser, w = Δp = 0

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Replacing:

0-0=h_{2} -h_{1} +\frac{V_{2}^{2}-V_{1}^{2}    }{2} +0

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Replacing:

0-0=h_{2} -400x10^{3}  +\frac{40^{2}-250^{2}      }{2} +0\\h_{2} =431430 J/kg=431.43kJ/kg

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m=\frac{P_{2} }{RT_{2} } A_{2} V_{2} \\\frac{6000}{3600} =\frac{100}{0.287*430} A_{2} *40\\A_{2} =0.051m^{2}

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