What is the lowest possible kinetic energy of the atom in

Question

Question.

Recently, a new element has been isolated, temporarily named Ununseptium, with atomic number 117. The longest-lived isotope obtained so far is ²⁹⁴Uus with an atomic mass of 294.2105 ± 0.0007u.

(a) How many electrons does a neutral atom of ²⁹⁴Uus contain?

(b) What is the mass n, of one ²⁹⁴Uus atom, in SI units? Include the error estimate in your answer, using 1 atomic mass unit lu = (1.66053892 ± 0.00000007) x 10^-27 kg.

Suppose that one ²⁹⁴Uus atom is confined within a particle detector, a one-dimensional box of length L = 200pm.

(c) What is the lowest possible kinetic energy of the atom in the box?

(d) If you measured the velocity of the atom while it is in the detector, what answer would you get?

(e) What velocity does the atom have in the absence of such a measurement?

(f) What is the total energy E, of the ²⁹⁴Uus atom?

²⁹⁴Uus is unstable, and decays rapidly into other atomic species. One possible decay route is via emission of an a-particle (a Helium nucleus, mass m_a = 4.0001u) to form element 115 (Uup), with atomic mass m_p = 290.1959u. Assume that the initial kinetic energy of the ²⁹⁴Uus atom can be neglected, so that the atom is approximately stationary in the lab frame.

(g) The total mass of the decay products is less than the mass of the original ²⁹⁴Uus atom. What accounts for this "missing mass" ?

(h) Conservation of total energy before and after the collision implies

E_s = E_a+ E_p

where E_a is the energy of the a particle, and E_p is the energy of the Uup atom. Similarly, conservation of total momentum before and after the collision implies

0 = P_a + P_p

Using these formulas, prove that the final energy of the a-particle is given by

E_a = m₂²+ m_a² - m_p²/ 2m_s c²

Hint: Solve for E_p, and use the mass-energy relation E² = c²p² + m²c⁴

(i) Hence calculate the energy and velocity of the a-particle.