Question

A. Calculate the energy in MeV released in a β+ decay of 22Na. Assume that the masses of 22Na and 22Ne are 21.994497 and 21.990439 u, respectively.

B.The laser system tested for inertial confinement can produce a 109.6 kJ pulse only 1.1 ns in duration. How many photons (in terms of 10²³ photons) are in the pulse if their wavelength is 0.97 μm?

C.Calculate the energy in MeV released in a α decay of ²⁴⁹Cf. Assume that the masses of ²⁴⁹Cf and ²⁴⁵Cm are 249.077455 and 245.060938 u, respectively.

D.Calculate the energy in MeV released in a β^- decay of ⁹⁰Sr. Assume that the masses of ⁹⁰Sr and ⁹⁰Y are 89.909088 and 89.90655 u, respectively.

E.The detail observable using a probe is limited by its wavelength. Calculate the energy in nanojoules of a γ-ray photon that has a wavelength of 2.84×10⁻¹⁶ m.

Answer 1

Solution:

Mass of ²²Na =21.994497u

Mass of ²² Ne=21.990439u

The difference in mass between the parent nucleus and products of the decay is $\bigtriangleup m$ =m(²²Na) -m(²²Ne)

Emitter Energy,$E=(\bigtriangleup m)c^{2}$

$(\bigtriangleup m)=21.994497u-21.990439u$

$(\bigtriangleup m)=0.004058u$

$E=(0.004058u)c^{2}$

$1u=931.5MeV/c^{2}$

Thus we obtain

$E=(0.004058)(931.5MeV/c^{2})(c^{2})$

$E=3.78MeV$