Q40 / 104 · NTA Abhyas (JEE Main) 2020 · Physics · Atomic Physics

An H-atom moving with speed v makes a head-on collision with another H-atom at rest. Both the atoms are in the ground state. The minimum value of velocity v for which one of the atoms may excite is

  • Correct answer
    A
    6.25×104 m s-1
  • B
    8×104 m s-1
  • C
    7.25×104 m s-1
  • D
    13.6×104 m s-1

Solution

For vmin collision should be completely inelastic.

According to energy conservation principle,

∴12mvmin2=12mv2+12mv2+∆E ...(i))

According to momentum conservation principle.

mvmin=mv+mv

∴v=mmin2 ...(ii)

After solving eq. (i) an d(ii), we get

12mvmin2=2∆E

∴vmin=4∆Em

Here, ∆E= minimum excitation energy

=10.2 eV

m=1.67×10-27 kg

∴vmin=4×10.2×1.6×10-191.67×10-27

=6.25×104 m s-1

Hence, the correct answer is option: 'A'.

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