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MODUL • Fizik TINGKATAN 5

SINAR KATOD (Menghuraikan) / CATHODE RAY (Describing)

Bergerak secara
garis lurus dalam
vakum
Moves linearly
in the vacuum

Mempunyai halaju,
tenaga kinetik dan Bercas
momentum yang tinggi negatif
Have high velocity, kinetic Ciri-ciri Negatively
energy and momentum sinar katod charged
The characteristics of
the cathode ray

Dipesongkan oleh
medan elektrik dan
medan magnet Menyebabkan skrin
berpendarflour
Deflected by the bercahaya
electric field and Cause fluorescent
magnetic field screen luminous

Contoh / Example

1 Beza keupayaan antara anod dan katod dalam senapang 2 Dalam tiub gambar set televisyen, beza keupayaan 20 kV
elektron ialah 5 kV. Hitungkan tenaga kinetik elektron. merentasi anod dan katod memecutkan sinar elektron. UNIT 4
[e = 1.6 × 10 C] Cas satu elektron ialah 1.6 × 10 C. Berapakah tenaga
–19
–19
UNIT 4
The potential difference between the anode and cathode in kinetik setiap elektron yang menghentam skrin?
an electron gun is 5 kV. Calculate the kinetic energy of the In a picture tube of a television set, a potential difference
–19
electrons. [e = 1.6 × 10 C] of 20 kV is applied across the anode and the cathode to
Penyelesaian / Solution accelerate the electron beam. The charge of each electron
–19
Tenaga keupayaan elektrik = Tenaga kinetik is 1.6 × 10 C. What is the kinetic energy of each electron
striking on the screen?
Electrical potential energy = Kinetic energy
eV = E Penyelesaian / Solution
3
(1.6 × 10 –19 C) × (5 × 10 V) = E E = eV –19 C) × (20 × 10 V)
= (1.6 × 10
3
E = 8.0 × 10 J = 32 × 10 –16
–16
= 3.2 × 10 J
–15
3 Dalam tiub vakum penerima televisyen, sinar katod Penyelesaian / Solution
dihasilkan dan memecut melalui beza keupayaan 7 kV. Tenaga keupayaan elektrik =Tenaga kinetik
Tentukan halaju sinar katod itu. Electrical potential energy = Kinetic energy
[e = 1.6 × 10 C dan m = 9 × 10 kg] 1 2
–19
–31
e
In the vacuum tube of a television receiver, a cathode ray eV = 2 mv
is produced and accelerated through a potential difference 1
3
7 kV. Determine the velocity of the cathode ray. 1.6 × 10 –19 C × 7 × 10 V = 2 × (9 × 10 –31 kg) × v 2
[e = 1.6 × 10 C and m = 9 × 10 kg] (1.6 × 10 –19 C)× (7 × 10 V)
–31
–19
3
e
v = (4.5 × 10 –31 kg)
2
v = 4.99 × 10 m s –1
7
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