An electron collides with a fixed hydrogen atom in its ground state. The 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 kinetic energy $(K.E.)$ of the colliding electron will be.....$eV$.

  • A
    $10.2$
  • B
    $1.9$
  • C
    $12.1$
  • D
    $13.6$

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Similar Questions

The binding energy of an electron in the ground state of a hydrogen atom is $13.6 \ eV$. What is the energy in $eV$ required to remove an electron from the first excited state of a $Li^{++}$ ion?

Energy of a stationary electron in the hydrogen atom is $E = -\frac{13.6}{n^2} \text{ eV}$. Calculate the energies required to excite the electron in a hydrogen atom to $(a)$ its second excited state and $(b)$ its ionized state,respectively.

What is an emission line?

In the above figure,$D$ and $E$ respectively represent:

The energy of a hydrogen atom in its ground state is $-13.6 \ eV$. The energy of the level corresponding to the quantum number $n = 2$ (first excited state) in the hydrogen atom is......$eV$.

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