Atomic Collision - Practice Questions & MCQ

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Quick Facts

Atomic Collision is considered one of the most asked concept.
4 Questions around this concept.

Solve by difficulty

Two hydrogen atoms in the ground state collide inelastically. The maximum amount by which their combined kinetic energy is reduced is:

10.2 eV

20.4 eV

13.6 eV

27.2 eV

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A H-atom moving with speed v makes a head on collision with a H-atom at rest. Both atoms are in ground state. The minimum value of velocity v for which one of the atom may excite is: $\left(\mathrm{m}_{\mathrm{H}}=1.67 \times 10^{-27} \mathrm{~kg}\right)$

$6.25 \times 10^4 \mathrm{~ms}^{-1}$

$8 \times 10^4 \mathrm{~ms}^{-1}$

$7.25 \times 10^4 \mathrm{~ms}^{-1}$

$13.6 \times 10^4 \mathrm{~ms}^{-1}$

View Solution

Concepts Covered - 1

Atomic Collision

Atomic Collision-

There are two ways to excite an electron in an atom-

1. By supplying energy to an electron through electromagnetic photons for eg., Photoelectric effect

2. By the atomic collision, the kinetic energy loss is utilized in the ionization or excitation of the atom.

Now let us understand the atomic collision -

Collision of a Neutron with an atom -

Let us consider an example of a head-on collision of a moving neutron with a stationary hydrogen atom as shown in the figure. Here. for
mathematical analysis, let us assume the masses of neutron and H atom to be the same -

Now there are two cases, the first is perfect elastic collision and another is a perfectly inelastic collision. Let us discuss these cases one by one -

1. Perfect elastic collision -

In this case, since the mass of the neutron and mass of the hydrogen atom, then the hydrogen atom will move with the same speed and kinetic energy which neutron is moving initially.

2. Perfect inelastic collision -

If both have perfect inelastic collision, then both move together. Now by applying conservation of momentum -

$\begin{array}{l}{m v_{0}=2 m v_{1}} \\ \\ \Rightarrow {v_{1}=\frac{v}{2}}\end{array}$

$v_o$ is the initial velocity of the neutron

$v_1$ is the final combined velocity of the atom and neutron.

Now the difference between initial and final kinetic energy is given as -

$\begin{aligned} \Delta E &=E_{i}-E_{f} \\ &=\frac{1}{2} m v_{o}^{2}-\frac{1}{2}(2 m)\left(\frac{v_{o}}{2}\right)^{2} \\ &=\frac{1}{2} m v_{o}^{2}-\frac{1}{4} m v_{o}^{2} \\ &=\frac{1}{4} m v_{o}^{2}=\frac{1}{2} E_{i} \end{aligned}$

Thus, half of the initial kinetic energy will be lost in the collision. The energy lost can only be absorbed by the atom involved in the collision and may get excited or ionized by this energy loss which takes place in case of inelastic collision. Here we are not considering the heat energy loss during the collision.

This loss in energy can be absorbed by the H atom only. From the previous concepts, we know that the minimum energy needed by the hydrogen atom to get excited is 10.2 eV for n =1 to n=2. So the minimum energy loss must be equal to 10.2 eV to excite hydrogen atoms. If the loss in energy is more than 10.2 eV then only 10.2 eV is absorbed by the hydrogen atom and the rest of the energy remains in the colliding particles (Neutron and H atom) as the collision is not perfectly inelastic.