When an element undergoes nuclear transmutation, the result is a completely different
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: -4.4 m/s
Explanation:
This problem can be solved by the Conservation of Momentum principle, which establishes that the initial momentum must be equal to the final momentum
:
(1)
Where:
(2)
(3)
is the mass of the child
is the initial velocity of the child
is the mass of the adult
is the initial velocity of the adult (it is sitting still)
is the final velocity of the child
is the final velocity of the adult
Substituting (2) and (3) in (1):
(4)
Isolating :
(5)
(6)
Finally:
This means the velocity of the child is in the opposite direction