<span>e=ca{\displaystyle e={\frac {c}{a}}}.</span>
Using the binding energy per nucleon number that you previously calculated for the isotope cobalt-60, is this a high amount of b
inding energy per nucleon according to the graph?
1 answer:
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Answer:
speed of electrons = 3.25 × m/s
acceleration in term g is 3.9 × g.
radius of circular orbit is 2.76 × m
Explanation:
given data
voltage = 3 kV
magnetic field = 0.66 T
solution
law of conservation of energy
PE = KE
qV = 0.5 × m × v²
v =
v =
v = 3.25 × m/s
and
magnetic force on particle movie in magnetic field
F = Bqv
ma = Bqv
a =
a =
a = 3.82 × m/s²
and acceleration in term g
a =
a = 3.9 × g
acceleration in term g is 3.9 × g.
and
electron moving in circular orbit has centripetal force
F =
Bqv =
r =
r =
r = 2.76 × m
radius of circular orbit is 2.76 × m