Question and Answer Health Contents

Question and Answer Health Occupation Contents Solution 1a 3 Solution 1b 3 Solution 2a 3 Solution 2bi 4 Solution 2bii 4 Solution 3ai 4 Solution 3aii 5 Solution 3b 5 Solution 3c 5 Solution 3d 5 Solution 4a 6 Solution 4b 6 Solution 4c 6 Solution 1a As given in question, Half life 2.72*365*24*60*60 = 85777920 seconds, Avogadro’s Number = 6.022 x 1023

Question and Answer
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Health

Question and Answer
Contents
Solution 1a.......................................................................................................................................3
Solution 1b.......................................................................................................................................3
Solution 2a.......................................................................................................................................3
Solution 2bi......................................................................................................................................4
Solution 2bii.....................................................................................................................................4
Solution 3ai......................................................................................................................................4
Solution 3aii.....................................................................................................................................5
Solution 3b.......................................................................................................................................5
Solution 3c.......................................................................................................................................5
Solution 3d.......................................................................................................................................5
Solution 4a.......................................................................................................................................6
Solution 4b.......................................................................................................................................6
Solution 4c.......................................................................................................................................6
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Question and Answer
Solution 1a
As given in question,
Half life T 1/ 2=¿2.72*365*24*60*60 = 85777920 seconds, Avogadro’s Number = 6.022 x 1023
Atomic weight = 273 gm,
Decay constant = 0.693/T1/2 = 0.693/ 85777920= 8.079 x 10-9 seconds
As given in question, 90% of material goes to ground state and 10 % excited state.
No of particle in excited state 2
273 x 6.022 x 1023 x 0.1 = 4.41172 x 1020 particle,
We know that, From question, No = 4.41172 1020, t = 20 seconds, λ=8.079 x 10−9
Nt =No e− λt Putting the value
Nt =4.41172 1020 x 2.72(−20∗8.079 x10−9 )
` Nt =4.41172 1020 x 2.72(−20∗8.079 x10−9 )
Nt =4.41172 x 1020
The gamma ray emitted from 2 gm of Isotopes X = 4.411 x 1020 rays
Solution 1b
The main characteristics of scintillation detectors are
 It must be solid and robust device, which is easily operated, the biased voltage required
for operation is quite low as compared to gas filled detectors.
 The scintillation detectors are moisture proof device. It does not require protection form
moisture.
 The energy lost by particle in scintillation detector should be quite low, as compared to
other devices.
 There should be fast response time, and able to detect minute energy fluctuation.
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Question and Answer
Solution 2a
The nucleus which is above the stability belt i.e. A in this case emits β particle, in this case
neutron is higher that proton in n/p ratio. The extra neutron, due to which nucleus is unstable
becomes stable after converting the neutron to proton, and places itself to stability belt to become
stable.
The nucleus which is below the stability lines generally emits α – particle, and losses two proton
and two neutrons. After alpha decay, the balance atoms become stable and shifted upward to
stable belt.
The nucleus around C, is table nucleus and it remains stable, at least for lower neutron. It dons
not emits any ray.
Solution 2bi
As given in question,
The relationship of Air karma and exposure is given as
Kair=33.97 ( μair
ρ )air
X
Putting the value =
Kair=33.97 ( 0.898
0.897 )air
air
=33.97 X Ans
Solution 2bii
It is the energy transferred per unit length of the track. In other words, it the quotient of
the dE/dl, where dE is average energy locally imparted to the medium by charged particle of
specified energy in traversing dl. That is L = dE/Dl.
The some of the example of high LET radiation is Neutrons, alpha particles. Whereas the
example of Low LET is x-rays and gamma particles.
Solution 3ai
The two-general body of UK legislation for Occupation and medical exposure of
ionisation energy is covered by Ionisation radiation regulation (IRR17) and radiation) emergency
preparedness and public information) regulation 2001, (REPPIR).
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Question and Answer
Solution 3aii
As per IRR17 regulation, the person responsible to control the Occupation and medical
exposure of ionisation energy is the employer, The employer may be self employed or any
designated person given by organisation,
Solution 3b
The minimum information required to X-ray examination of a patient is related with following
condition,
 The patient should be uniquely identified.
 The medical exposure limit should be justified with available clinical information about
the patient.
 The pregnancy information of last menstrual information is necessary.
Solution 3c
Effective dose,
As per IRR17, the effective dose is related to overall long-term risk for a person by doing a
ionisation procedure, It helps compare risks from different procedure.
Equivalent dose
As per IRR17, equivalent does is the amount of exposure of ionisation energy, that takes
damaging property of different type of radiation into account. Further it is given that not all
radiation given the same exposure. The radiation for diagnostic purposes,
For abdomen area, the equivalent does is 20 mSv, and effective dose is 15 mSv.
Solution 3d
The six measure that is generally being taken for staff protection is
 Architectural wall shielding
 Table suspended drapes
 Using lead aprons
 Use of thyroid collars
 Use of glass with lead
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Question and Answer
 Separating the stand of shield and drapes
Solution 4a
There are several processes in which excited molecules are stabilised and its photon emission is
taking place.
 Vibrational relaxation: In this process, when two excited molecules strike each other,
then there is slight increase in temperature, leads to extra heat generation between
molecules. This molecule releases energy into photon and return back to ground state.
This process is so fast that, molecule can terminates at any ground state condition.
 Phosphorescence: When the transition of molecules through intersystem crossing the
triplet state. The deactivation process occurs through fluorescence or phosphorescence’s
with photon emission.
Solution 4b
In laser system, the population inversion is achieved by using three metastable levels.
This is due to the reason that, excitation of electron is between two levels, but electron in upper
energy level goes down by releasing the energy into metastable level. In this condition, the upper
level always remains empty i.e. practically unpopulated most of the time. This excitation process
is very frequent used in laser transition, further itis not triggering stimulated emission.
In the given figure we can see that, electron going up to second level and by releasing energy; it
comes back to ground level. In this condition upper level is almost empty.
Solution 4c
The hazard that associated with class IV laser includes, biological hazard either direct or diffused
reflection. These are the following which comes under class iv laser hazard
 Ocular injury
 Tissue damage
 Respiratory hazard, and
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Question and Answer
 Fire and explosion,
Solution 5
As given in question
Nt =No e− λt Suppose No = 1000 at 02:05 time
Nt = 400 at 02:10 time
Then putting the value in equation,
Nt =No e− λx300…………(i)
Or 400/1000 ¿ 2.71− λx 300
Taking log on both sides
Ln0.4 = −λx 300 x ln2.71
Ln0.4-ln2.71 = −λx 300
−λx 300 =- 0.83724
Or λ= 0.83724/300 = 0.002791
Now λ = 0.002791, Nt = 1000, No =?, t = 300 seconds
Again, putting the value in equation (i)
1000/ No =e−0.83724
Or, No = 1000
2.71−0.83724 = 1000
0.432964 =2309.659
Then value of total gamma ray flux at 2:00 O clock = 2309.659 per second
Solution 6
As given in question,
The photon energy α=5.11 keV
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Question and Answer
The photon energy α=5.11 MeV
From Crompton scattering energy, we know that
E '= E
1+ α(1−cosθ) and, α = E
mo c2
We know that maximum energy transfer when angle is 180o
Then, α = 5.11 x 103
511 x 103 = 0.01
, α =¿ 0.01
Now Emax at 5.11 Kev
E '= 5.11 x 103
1+0.01( 1−cos 180o )= 5.11 x 103
1+ 0.01(1+1)= 5.11 x 103
1.002 =5089.82=5.08982 KeV
Now Emax at 5.11 Mev
E '= 5.11 x 105
1+0.01( 1−cos 180o )= 5.11 x 105
1+ 0.01(1+1)= 5.11 x 105
1.002 =5089.82=5.08982 M eV
Ans
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Question and Answer
References
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