WAEC 2018 PHYSICS THEORY AND OBJECTIVE ANSWER / EXPO / RUNZ





Physics OBJ:
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4a)
An intrinsic semiconductor is an undoped semiconductor that is a pure semiconductor without any significant dopant species present.

4b)
The P Type semiconductor is a type of semiconductor that carries a positive charge, while the N type semiconductor carries a negative charge.

5)
Range = u²Sin2tita/g
At maximum range
Sin2tita = 1
2tita =sin^-1(1)
2tita = 90dgrees
Tita = 90/2 = 45degree
Maximum height reached = u²sin²tita/2g
=u²(sin45)²/2g
=200²(sin45)²/2(10)
=40000(1/√2)2/20
=40000(1/2)/20
=20000/20
=1000metres..

6)
Given constant = 2.9×10^-3mk
Temperature = 57degreeC = (57+273)k = 330k
Using landamaxT = constant
landamaxT330 = 2.9×10^-3
landamax = 2.9×10^-3/330
landamax = 8.788×10^-6m
The speed of electromagnetic wave, v = 3×10^8m/s
Using V = f landa
f = v/landa
=3×10^8/8.788×10^-6

(12a) binding energy is the minimum energy that would be required to disassemble the nucleus of an atom into its component parts. These component parts are neutrons and protons, which are collectively called nucleons.
(12b) They have short wire length and high frequency. -They are highly penetrating -They travel in straight line -They don’t require material medium for their propagation.

(12c)
-It is used in production of electricity.
-It is used to study and detect charges in genetic engineering.
-It is used in agriculture.
-It is used in treatment of cancer.

(12di)
E = hf-hfo
but f = v/landa
E= v/landa.h – wo
Where wo = hfo = work function
f= frequency
landa = wavelength

Hence
hf = hfo – E
f = hfo – E/h
f = wo – E/h
Recall; that v = f landa
Therefore f = v/landa = 3×10^8/4.5×10-7
=3/4.5 × 10^8+7
=6.6×10^14Hz
f = 6.6×10^14Hz

(12dii)
E = hf
=6.6×10^-34 × 6.6×10^14Hz
=43.56×10^-20J

(12diii)
Energy of the photoelectron E = hf – vo
=Energy of incident electron – work function
=4.356×10^-19J – 3.0×10^-19J
=1.356×10^-19J

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