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A satellite in orbit could derive its energy from a very strong radioactive source of alpha particles. Suppose that 30 W (two significant digits) of electrical power are to be produced with 5.0% efficiency by conversion of the (kinetic) energy of the alpha particles. If the energy of each alpha particle is 5.5 MeV, how many particles must be emitted by the source per second?

Respuesta :

Manetho

Answer:

6.23×10^(14) particles

Explanation:

Let the total power be P_0

now according to question

5% ×P_0=30W

so P_0=30/0,05=600 W.

an alpha particle'energy is 5.5 MeV

Let n be number of alpha particles emitted per sec.

Therefore,

n=600J/5.5MeV=600/(5.5×e^6×1.6×e^-19)=6.82×e^14(particles)

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