whatIs the mihimum KE the electron heeds to escape the He nucleus (in J and ev)?

Question

whatIs the mihimum KE the electron heeds to escape the He nucleus (in J and ev)? 3 (25.58). In the form of radioactive decay known as alpha decay, an unstable nucleus emits a helium-atom nucleus, which is called an alpha particle. An alpha particle contains two protons and two neutrons (no electrons). Suppose a uranium nucleus with 92 protons decays into thorium, (90 protons) and an alpha particle: U Th + He Our problem begins with the alpha particle initially at rest at the surface of the thorium nucleus which is 15 fim in diameter. What is the speed of the alpha particle when it is detected in the laboratory? 4 (23.48). See next page

Explanation / Answer

According to the concept of the nuclear physics

K. E=9*10^9*qQ/r

Charge q=particle =3.2*10^-19 C

Total charge Q=90*1.6*10^-19=144*10^-19 C

Diameter D=15*10^-15 m

Radius r=7.5*10^-15 m

Mass of the alpha particle m=6.64*10^-27 kg

Now we find the kinetic energy of alpha particle

KE=9*10^9*3.2*10^-19*144*10^-19/(7.5*10^-15)

=552.96 *10^-14 J

Now we find the speed of the alpha particle

1/2mv^2=552.96*10^-14

v^2=(2*552.96*10^-14)/6.64*10^-27

=166.6*10^-13

=16.66*10^-12

Speed v=4.082*10^-6 m/s

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