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Ganezh [65]
3 years ago
8

An X-ray tube emits X-rays with a wavelength of 1.0 x 10-11 m. Calculate the potential that must be applied across the X-ray tub

e to give each incident electron its energy, in volts.
A ) 0.6 x 105
B ) 1.2 x 105
C ) 2.4 x 105
D ) cannot be determined
Physics
2 answers:
sergey [27]3 years ago
3 0

Answer:

1.24 x 10 to the 5 ev = 124,000 ev          its B

Explanation:

E = hc/lambda = 1.24 ev-micrometer/1.0x10 to the -5 micrometers = 1.24 x 10 to the 5 ev = 124,000 ev

h = Planck's constant = 6.626 × 10 to the -34 joule·s

c = speed of light = 2.998 × 10 to the 8 m/s

lambda is the given wavelength

E is the desired photon energy

mel-nik [20]3 years ago
3 0

The potential that must be applied across the X-ray tube to give each incident electron the appropriate energy to goose X-rays with a wavelength of 1.0 x 10-11 m out of the anode cannot be determined from the given information. <em>(D)</em>

The energy (or wavelength) of the emitted X-rays is NOT directly determined by the energy of the incident electrons in the x-ray tube. It's determined mainly by the substance of which the anode is composed, and the electron configuration of the substance.

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1 year ago
A cylindrical resistor element on a circuit board dissipates 1.2 W of power. The resistor is 2 cm long, and has a diameter of 0.
34kurt

Answer:

(a) The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) The fraction of heat dissipated from the top and bottom surfaces is 0.045.

Explanation:

(a) The amount of heat dissipated (Q), measured in joules, by the cylindrical resistor is the power multiplied by operation time (\Delta t), measured in hours. That is:

Q = \dot Q \cdot \Delta t (1)

If we know that \dot Q = 1.2\,W and \Delta t = 86400\,s, then the amount of heat dissipated by the resistor is:

Q = (1.2\,W)\cdot (86400\,s)

Q = 103680\,J

The resistor disspates 103680 joules during a 24-hour period.

(b) The heat flux (Q'), measured in watts per square meter, is the heat transfer rate divided by the area of the cylinder (A), measured in square meters:

Q' = \frac{\dot Q}{A} (2)

Q' = \frac{\dot Q}{\frac{\pi}{2}\cdot D^{2}+\pi\cdot D \cdot h } (3)

Where:

D - Diameter, measured in meters.

h - Length, measured in meters.

If we know that \dot Q = 1.2\,W, D = 4\times 10^{-3}\,m and h = 2\times 10^{-2}\,m, the heat flux of the resistor is:

Q' = \frac{1.2\,W}{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2}+\pi\cdot (4\times 10^{-3}\,m)\cdot (2\times 10^{-2}\,m) }

Q' \approx 4340.589\,\frac{W}{m^{2}}

The heat flux of the resistor is approximately 4340.589 watts per square meter.

(c) Since heat is uniformly transfered, then the fraction of heat dissipated from the top and bottom surfaces (r), no unit, is the ratio of the top and bottom surfaces to total surface:

r = \frac{\frac{\pi}{2}\cdot D^{2}}{A} (3)

If we know that A \approx 2.765\times 10^{-4}\,m^{2} and D = 4\times 10^{-3}\,m, then the fraction is:

r = \frac{\frac{\pi}{2}\cdot (4\times 10^{-3}\,m)^{2} }{2.765\times 10^{-4}\,m^{2}}

r = 0.045

The fraction of heat dissipated from the top and bottom surfaces is 0.045.

7 0
3 years ago
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