Calculate the nuclear binding energy per nucleon for 136^Ba if its nuclear mass is 135.905 amu.
1 answer:
Answer:
1.312 x 10⁻¹² J/nucleon
Explanation:
mass of ¹³⁶Ba = 135.905 amu
¹³⁶Ba contain 56 proton and 80 neutron
mass of proton = 1.00728 amu
mass of neutron = 1.00867 amu
mass of ¹³⁶Ba = 56 x 1.00728 amu + 80 x 1.00867 amu
= 137.10128 amu
mass defect = 137.10128 - 135.905
= 1.19628 amu
mass defect = 1.19628 x 1.66 x 10⁻²⁷ Kg
= 1.9858 x 10⁻²⁷ Kg
speed of light = 3 x 10⁸ m/s
binding energy,
E = mass defect x c²
E = 1.9858 x 10⁻²⁷ x (3 x 10⁸)²
E = 17.87 x 10⁻¹¹ J/atom
now,
binding energy per nucleon =
= 0.1312 x 10⁻¹¹ J/nucleon
= 1.312 x 10⁻¹² J/nucleon
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Answer:
The position of the car at t = 1.5 s is at -8.1625 meters
Explanation:
The initial position of the car is 3.2 meters
The initial velocity is -8.4 m/s
The constant acceleration is 1.1 m/s²
We need to find the final position of the car at the time t = 1.5 seconds
The displacement <em>s</em> = final position - initial position
, where <em>u</em> is the initial velocity, <em>a</em> is the
constant acceleration and <em>t</em> is the time
So we can find the final velocity by using the rule:
final position - initial position =
initial position = 3.2 meters , u = -8.4 m/s , a = 1.1 ²m/s , t = 1.5 s
Substitute these values in the rule
final position - 3.2 =
final position - 3.2 = -12.6 + 1.2375
final position - 3.2 = -11.3625
add 3.2 for both sides
final position = -8.1625
<em>That means the car is at 8.1625 meters in opposite direction</em>
<em>The position of the car at t = 1.5 s is at -8.1625 meters </em>