Question

A hydrogen atom moving at speed υ collides with another hydrogen atom kept at rest. Find the minimum value of υ for which one of the atoms may get ionized.
The mass of a hydrogen atom = 1.67 × 10−27 kg.

Answer 1

The minimum value of v is $7.2 \times 10^8$ for which one of the atoms get ionized.

Explanation:

It is given that

Hydrogen atom’s mass, M = 1.67 × 10−27 kg

Let the velocity be v with which the atom of the hydrogen is traversing before collision.

After the collision, let the velocities of the hydrogen atoms be  $v_1$ and $v_2$.

For the ionization of one hydrogen atom, the energy used,

$\Delta E = 13.6\times (1.6 \times 10^{-19} ) J$

When the momentum conservation is applied, we obtain

$mv = mv_1 + mv_2$…(1)

When the conservation of mechanical energy is applied, we obtain

$12mv_2 = 12mv_2^{2} + \Delta E + 12mv_1^{2}$               …(2)

When equation (1) is used, we obtain

$v_2 = v_2^{2} + v_1^{2} + 2v_1v_2$          …(3)

When both the sides of equation (2) is multiplied by 2 and divided by m.

$v_2 = v_1^{2} + v_1^{2} + 2\frac{\Delta E}{m}$             ….4

When (4) with compared with (3), we obtain

Therefore, $2v_1v_2 = (v_1 + v_2)^{2} - 4 v_1 v_2$

So, $v_1 - v_2^{2} = v_2 - \frac{4 \Delta E}{m}$

For v’s minimum value,

$v_1 = 0 = v_2$

Also,

$v_2 - \frac{4\Delta E}{m} = 0$

Therefore, $v_1 - v_2^{2} = v_2 - \frac{4\Delta E}{m}$  can be written as

$v_2^{2} = \frac{4 \Delta E}{m}$

$v_(min) = \sqrt{\frac{4 \Delta E}{m}} = v_2$

$\begin{aligned}&v_{\min }=\sqrt{\frac{4 \times 13.6 \times 16 \times 10^{-19}}{167 \times 10^{-27}}}\\&=7.2 \times 10^{8} \mathrm{m} / \mathrm{s}\end{aligned}$                          

The minimum value of v is $7.2 \times 10^8$ meter per second for which one of the atoms may get ionized.

Answer 2

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