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Two hydrogen atom in ground state are mo...

Two hydrogen atom in ground state are moving on opposite direction with the same speed and collide head on . The minimum kinetic energy of each hydrogen atom for the collision to be inelastic so that both the atoms are excited is

A

`13.6eV`

B

`10.2eV`

C

`20.4eV`

D

`27.2eV`

Text Solution

Verified by Experts

The correct Answer is:
B
Let K by the initial energy of each atoms . Maximum loss in Ke takes place in perfectly inelastic collision. From conservation of linear momentum `"mu"+(-"mu")=2mvRightarrowv=0`
Where v=common velocity after collision.
therefore maximum possible loss in KE is
` DeltaK=1/2"mu"^(2)xx2-1/22mxx0^(0)=21/2"mu"^(2)=2K` Minimum energy required to excite both the electrons is `DeltaE=10.2+10.2=20.4eV`. For inelastic collision with two electron excited
`DeltageDeltaERightarrow2Kge20.4RightarrowK_(min)=10.2eV`
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Knowledge Check

  • The ionization energy of hydrogen atom in the ground state is

    A
    13.6 MeV
    B
    13.6 eV
    C
    13.6 Joule
    D
    Zero

  • The minimum enegry required to excite a hydrogen atom from its ground state is

    A
    `13.6 eV`
    B
    `-13.6 eV`
    C
    `3.4 eV`
    D
    `10.2 eV`

  • The minimum energy required to excite a hydrogen atom from its ground state is .

    A
    13.6 eV
    B
    10.2 eV
    C
    3.4 eV
    D
    1.51eV

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