Chapter 9: Q121P (page 256)

An electron undergoes a one-dimensional elastic collision with an initially stationary hydrogen atom. What percentage of the electron’s initial kinetic energy is transferred to kinetic energy of the hydrogen atom? (The mass of the hydrogen atom is 1840 times the mass of the electron)

Short Answer

Expert verified

The percentage of the electron’s initial kinetic energy transferred to kinetic energy of the hydrogen’s atom is 0.22%.

Step by step solution

Step 1: Given

The mass of the hydrogen in terms of the mass of the electron is $m_{H} = 1840 m_{e}$

Determine the formula for the velocity and kinetic energy:

Using the formula for velocity of the target object in an elastic collision in one dimension, find the relation between final velocity of hydrogen and initial velocity of the electron. Using this relation, write the equations for their kinetic energy and can find the percentage of the electron’s initial kinetic energy transferred to kinetic energy of the hydrogen’s atom.

Formulae are as follow:

i) $v_{2 f} = \frac{2 m_{1}}{m_{1} + m_{2}} v_{1 i}$

ii) $K E = \frac{1}{2} m v^{2}$

Here, m, m₁, m₂ are masses, v, v₁, v₂ are velocities and KE is kinetic energy.

Determine the percentage of the electron’s initial kinetic energy transferred to kinetic energy of the hydrogen’s atom

For an elastic collision between the hydrogen atom and the electron,

$V_{f H} = \frac{2 m_{e}}{m_{e} + m_{H}} V_{i e}$

$V_{f H} = \frac{2 m_{e}}{m_{e} + 1840 m_{e}} V_{i e}$

$V_{f H} = \frac{2 m_{2}}{1841 m_{e}} v_{i e}$

$V_{f H} = \frac{2}{1841} V_{i e}$

Now, for kinetic energy:

$K E = \frac{1}{2} m v^{2}$

Therefore, kinetic energy of electron and hydrogen atom is:

$K E_{i e} = \frac{1}{2} m_{e} V_{i e}^{2}$

$K E_{f H} = \frac{1}{2} m_{H} V_{f H}^{2}$

Substitute the values and solve as:

$K E_{f H} = \frac{1}{2} (1840 m_{e}) \frac{2^{2}}{{(1841)}^{2}} v_{i e}^{2}$

$K E_{f H} = \frac{1840 \times 4}{{(1841)}^{2}} \times \frac{1}{2} m_{e} V_{i e}^{2}$

$K E_{f H} = \frac{1840 \times 4}{{(1841)}^{2}} K E_{i e}$

$\frac{K E_{f H}}{K E_{i e}} \times 100 = \frac{1840 \times 4}{{(1841)}^{2}} \times 100$

Solve further as:

$\frac{K E_{f H}}{K E_{i e}} \times 100 = 0.217 % = 0.22 %$

Hence, the percentage of the electron’s initial kinetic energy transferred to kinetic energy of the hydrogen’s atom is 0.22%.

Therefore, the percentage of the electron’s initial kinetic energy transferred to the kinetic energy of the hydrogen’s atom can be found using the formula for velocity of the target object in an elastic collision in one dimension.