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madam [21]
2 years ago
5

Briefly discuss:

Physics
1 answer:
IrinaK [193]2 years ago
8 0

Answer:

(i) Si device are coated with B or Li for neutron detection

(ii) Fast neutrons are normally first slowed down  to thermal energies before measurement

(iii) The strip detectors consists of p⁺ and n implants in the region of a very thin (300 to 400 μm) depletion zone  though which the neutrons traverse producing a readout based on the generated charges being directed to the cathode, p⁺, material  which then transmits the charge to the device electronics

Explanation:

(i) Detection of neurons with only an Si device in not possible due to the large neutrons path in Si such that the silicon needs to be coated with B or Li which readily interact with neutrons. The neutron interaction with the reactive coating produces an alpha particle which can be detected by the semi conductor and a nucleus

(ii) Neutrons having a kinetic energy that is more than 1 MeV which as such has a velocity of more than 15,000 km/s is known as fast neutrons or fission neutrons. The fast neutrons are slowed in a nuclear reactor by neutron moderation to thermal energies

Due their high speed, fast neutrons are normally slowed down which however results in the loss of some vector properties of the neutron

Techniques for fast neutron detection includes,

1) Recoil detectors which are capable of fast neutron detection without moderation

2) Bonner spheres detector first converts the fast moving electron to slow down before detection

3) Scintillation counter are widely used but require the conversion of the neutron to a charged particles before detection

(iii) Strip detectors provide high precision measurement of a particle's crossing point which can be further improved by use of low noise electronics

Applications of silicon strip detectors include

1) Particle tracking in research in particle physics

2) Particle tracking in researches in x-spectroscopy nuclear research

3) Imaging in x-talography

4) Medical research imaging

5) Particle tracking and imaging in astrophysics.

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Katarina [22]

Answer:

Frequency of oscillation, f = 4 Hz

time period, T = 0.25 s

Angular frequency, \omega = 25.13 rad/s

Given:

Time taken to make one oscillation, T = 0.25 s

Solution:

Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:

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Therefore, time period, T = 0.25 s

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\omega = 2\pi \times 4

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