Tamil Nadu Board of Secondary EducationHSC Science Class 12th

Consider two hydrogen atoms HA and HB in ground state. Assume that hydrogen atom HA is at rest and hydrogen atom HB is moving with a speed and make head-on collide on the stationary hydrogen atom HA. - Physics

Numerical

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

Solution

Collision between hydrogen HA and hydrogen HB atom will be inelastic if a part of kinetic energy is used to excite atom.

If u1 and u2 are speed of HA and HB atom after collision, then

mu = mu1 + mu2 …… (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 10.2eV

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

The minimum K.E of the moving hydrogen atom HB is 20.4 eV.

Concept: Electric Discharge Through Gases
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Tamil Nadu Board Samacheer Kalvi Class 12th Physics Volume 1 and 2 Answers Guide
Chapter 9 Atomic and Nuclear physics
Evaluation | Q 1. | Page 191
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