When exposed to a radioactive source which emits 1.2-MeV gamma-rays, a particular material is found to have a half-value thickne
1 answer:
Answer:E
Explanation:
It is given that Energy of gamma ray is E=1.2 Mev
Shielding effect can be measured by measuring the fraction of gamma rays blocked by shield. If certain thickness will able to block half the radiation then to block 75% radiation we need to add same amount of thickness in order to block the remaining radiation.
i.e. fraction is blocked by 10 cm thickness
then remaining radiation is
another 10 cm thickness will block the remaining half radiation i.e.
so total 75 % radiation will be blocked
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Answer:
(a) 5.43 x 10⁵ J
(b) 3.07 x 10⁵ J
(c) 45 °C
Explanation:
(a)
= Latent heat of fusion of ice to water = 3.33 x 10⁵ J/kg
m = mass of ice = 1.63 kg
= Energy required to melt the ice
Energy required to melt the ice is given as
= m
= (1.63) (3.33 x 10⁵)
= 5.43 x 10⁵ J
(b)
E = Total energy transferred = 8.50 x 10⁵ J
Q = Amount of energy remaining to raise the temperature
Using conservation of energy
E = + Q
8.50 x 10⁵ = 5.43 x 10⁵ + Q
Q = 3.07 x 10⁵ J
(c)
T₀ = initial temperature = 0°C
T = Final temperature
m = mass of water = 1.63 kg
c = specific heat of water = 4186 J/(kg °C)
Q = Amount of energy to raise the temperature of water = 3.07 x 10⁵ J
Using the equation
Q = m c (T - T₀)
3.07 x 10⁵ = (1.63) (4186) (T - 0)
T = 45 °C