Tutorial 1

RADIATION SCIENCE Tutorial 1 1. Find the answer of the fraction below

A ) 3 + 2 – 1 = 81 7 5 4 140 B) 2 x 2 ÷ 2 = 6 7 5 3 35 C ) 2 + 1 x 3 ÷ 1 - 2 = 109 3 4 7 3 5 140 D ) 11 x 31 + 135 ÷ 163 - 324 = 16 47 238 235 431 2. Write into decimal point at 3 decimal places A ) 31 = 0.660 47 B ) 133 = 0.561 237 C ) 163 = 0.703 232 3. Find the answer of A if given x = 3 , y = 4 and z = 5 A) x = z Y A

A = 6.667

B ) 2x - 3y = 3A - z

A = -1 3

C ) x = zAy

A = 0.880 ( using log )

D ) x = yzA

A = - 0.179 ( using log )

E ) A = xe-Ay

A = 6.184 x 10-9

F ) x = ye-Az A = ln

yx

z A = 0.0575

4. Find the answer below A ) 2.5 km x 3.2 mm - 55.36 cm = 7.4464 m B ) 3.4 Bq ÷ 2.1 mBq x 0.39 KBq = 5.18 x 105 Bq 5.

Write all the radiological unit of radiation and quality in SI units

1. 2. 3. 4.

Quantity Exposure Absorbed Dose Effective Dose Radioactive

Customary Unit Name Symbol Roentgen R rad Rad Rem Rem Curie Ci

SI Unit Name Symbol Air kerma Gya Gray Gyt seivert Sv bacqueral Bq

6. Write the answer in SI units A ) 250 Roentgen = 2.5 Gya B ) 345 Rad = 3.45 Gyt C ) 53 rem = 0.53 sv D ) 53 µCi = 1.961 x 103 Bq 7.

Write the history who discover the radioactive element of alpha ( beta ( β ) and gamma ( γ )

α

),

Ernest Rutherford was the person who discover the radioactive element of alpha ( α ) , beta ( β ) and gamma ( γ ). A beam of positive charge alpha particle was projected againt a thin metal foil. Most of the alpha particle past through the foil as it is empty sphere but a few particle deflected from their original direction of travel were scathered through large angle. Some particle even deflected backwarel, reversing their direction of travel. Rutherford explained this astounding result by assuming that the positive charge is an atom was concentrated in a region that was small relative to the size call nucleus of atom. 8.

Using the general formulated are modified by Bork ( medification formulated ) A ) Write the formula

En = z2 ( 13.6 ev ) n2 rn = n2 ( 0.0529 ) z B ) Find the energy (En) and radius (rn) of atomic helium En = rn =

41 12

( 13.6 ev ) = 54.4 ev ( 0.0529 nm ) = 0.0132 nm

9. Find the number of proton, electron and nucleon A)

126

C

Proton = Electron = Neutron = Nucleon = B)

2713

Al

Proton = Electron = Neutron = Nucleon =

C)

13756

18474

13 13 14 27

Ba

Proton = Electron = Neutron = Nucleon = D)

6 6 6 12

56 56 81 137

W ( Tungsten )

Proton = 74 Electron = 74

42 H

Neutron = 110 Nucleon = 184 E)

23492

u

Proton = Electron = Neutron = Nucleon =

92 92 146 234