Consider two hydrogen atoms H_{A} and H_{B} in ground state. Assume that hydrogen atom H_{A} is at rest and hydrogen atom H_{B} is moving with a speed and make head-on collide on the stationary hydrogen atom H_{A}. After the strike, both of them move together. What is the minimum value of the kinetic energy of the moving hydrogen atom H_{B}, such that any one of the hydrogen atoms reaches one of the excitation state?

#### Solution

Collision between hydrogen H_{A} and hydrogen H_{B} atom will be inelastic if a part of kinetic energy is used to excite atom.

If u_{1} and u_{2} are speed of H_{A} and H_{B} atom after collision, then

mu = mu_{1} + mu_{2} …… (1)

`1/2 "mu"^2 = 1/2 "mu"_1^2 + 1/2 "mu"_2^2 + Delta "E"` ....(2)

`"u"^2 = "u"_1^2 + ("u" - "u"_1)^2 + (2Delta"E")/"m"`

`"u"_1^2 - "uu"_1 + (2Delta"E")/"m" = 0`

For `"u"_1` to be real

`"u"^2 - (4Delta"E")/"m" >= 0`

`"mu"^2/2 >= 2 xx Delta "E"`

Δ E = 10.2 eV

Thus `(1/2 "mu"^2)_"min" = 2 xx 10.2`eV

`(1/2 "mu"^2)_"min" = 20.4` eV

The minimum K.E of the moving hydrogen atom H_{B} is 20.4 eV.