The first step of the scientific method is
2 answers:
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Answer:
the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Explanation:
Given that;
diameter D = 2.0 mm
current I = 1.0 mA
K.E of each proton is 20 MeV
the number density of the protons in the beam = ?
Now, we make use of the relation between current and drift velocity
I = MeAv ⇒ 1 / eAv
The kinetic energy of protons is given by;
K = v²
v = √( 2K / )
lets relate the cross-sectional area A of the beam to its diameter D;
A = πD²
now, we substitute for v and A
n = I / πeD² ×√( 2K /
)
n = 4I/π eD² × √( / 2K )
so we plug in our values;
n = ((4×1.0 mA)/(π(1.602×10⁻¹⁹C)(2mm)²) × √(1.673×10⁻²⁷kg / 2×( 20 MeV)(1.602×10⁻¹⁹ J/ev )
n = 1.98695 × 10¹⁸ × 1.6157967 × 10⁻⁵
n = 3.2 × 10¹³ m⁻³
Therefore, the number density of the protons in the beam is 3.2 × 10¹³ m⁻³
Answer:
λ =8.57 μ m
Explanation:
Given that
Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C
Standard form
Ey=Eo cos[k x-ωt] N/C
By comparing the given equation with the standard wave equation
Eo = 375 N/C
ω = 2.20 × 10¹⁴ rad/s
We know that ω = 2 π f
f=3.50×10¹³ Hz
We know that the velocity given as
V = f λ
λ =Wavelength
V=Speed = 3 x 10⁸ m/s
λ =0.00000857 m ( 1 μ m = 10⁶ m)
λ =8.57 μ m