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An electron collides with a fixed hydrog...

An electron collides with a fixed hydrogen atom in its ground state. Hydrogen atom gets excited and the colliding electron loses all its kinetic energy. Consequently the hydrogen atom may emit a photon corresponding to the largest wavelength of the Balmer series. The minimum K.E. of colliding electron will be

A

10.2 eV

B

1.9 eV

C

12.1 eV

D

13.6 eV

Text Solution

Verified by Experts

Excitation up to `n =3` is required so that visible length is emitted upon de-excitation.
`E _(1) = ( - 13.6)/(1 ^(2)) , E _(2) = (-13. 6)/(9)therefore E _(2) - E_(1) = (-13.6)/(9) + (13.6)/(1)`
So required energy `=13.6 (1 - (1)/(9)) = 12. 1 eV`
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Knowledge Check

  • The ratio of the total energy E of the electron to its kinetic energy K in hydrogen atom is

    A
    1
    B
    `(1)/(2)`
    C
    2
    D
    `-1`
  • Hydrogen atom is said to be in its ground state when its orbiting electron

    A
    has left the atom
    B
    is at rest
    C
    has the lowest energy state
    D
    has fallen to the nucleus
  • Energy of the hydrogen atom in its ground state is -13.6eV the energy corresponding to the second excited state is

    A
    `-1.51eV`
    B
    `10.2eV`
    C
    13eV
    D
    `-3.4eV`
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