Question

A nitrogen atom of m_{1}=14 u is moving with an initial
kinetic energy K_{i} and strikes a stationary tungsten atom
of m_{2}=184 u. During the collision, some of the kinetic
energy of the incident atom is used to ionize the stationary atom;
an energy E is needed to eject one electron. The ejected electron
has less than 0.0001 times the mass of the atoms, so it carries
away a negligible momentum. After the collision, the two atoms
stick together and move off as a diatomic ionized molecule with
kinetic energy, K_{f}. *Eventually* you will be
asked to find the minimum kinetic energy of the incident atom
needed to ionize the stationary atom.

**20.** [1pt]

Select the statements that are true; e.g., if only statements A and
B are true, enter AB; if none are true, enter N. *You only have
3 tries!*

Our professor is crazy to assign such a problem because we have not been taught the material needed to complete it.

If the (initially stationary) atom becomes ionized, then
K_{i} ≥ K_{f}+E.

The energy used in the ionization process is analogous to energy stored in a form other than kinetic energy (e.g., as in a compressed spring).

During the collision the linear momentum is conserved.

**Answer:**

Answer #1

**let m1 = 14*u
m2 = 184*u
let vi is the initial speed of m1.**

**let vf is the final speed of the combined
molecule.**

**Apply conservation of momentum**

**final momentum = initial momentum**

**(m1 + m2)*vf = m1*vi**

**(14*u + 184*u)*vf = 14*u*vi**

**vf = vi*14/(14 + 184)**

**vf = 0.0707*vi**

**Ki/Kf = (1/2)*m1*vi^2/((1/2)*(m1 +
m2)*vf^2)**

**= 14*vi^2/((14 + 184)*(0.0707*vi)^2)**

**= 14/(198*0.0707^2)**

**= 14.1**

**Ki = 14.1*Kf**

**20)**

**If the (initially stationary) atom becomes ionized, then Ki
≥ Kf+E. ---> True**

**The energy used in the ionization process is analogous
to energy stored in a form other than kinetic energy (e.g., as in a
compressed spring). ---> False**

**During the collision the linear momentum is conserved.
--> True**

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