What minimum amount of energy is required to bring an electron from ground state of be 3+ to infinity?

Option 3 : ionization energy

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CONCEPT:

• The minimum energy required to remove an electron from the ground state to outside the atom is called ionization energy.

The energy of electrons in any orbit is given by:

\({\bf{Energy}}\;{\bf{in}}\;{\bf{nth}}\;{\bf{orbit}}{\rm{\;}}\left( {{E_{n\;}}} \right) = \; - 13.6\;\frac{{{Z^2}}}{{{n^2}}}\;eV\)

Where n = principal quantum number and Z = the atomic number

• This is the energy that is required to remove an electron from its orbit.

CALCULATION:

• The minimum energy that is given to the electron and leaves the atom, this energy is known as the ionization energy.
• So the correct answer is option 3.

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"Where would this energy go?" Several places.

"Would it get converted into radiation due to the electron's accelerating towards the proton," Most of the energy would go into this channel, at least that's what experiments would suggest. The cascade could produce anything from a single Lyman limit photon, to a Balmer limit photon plus a Lyman alpha photon, a Lyman beta photon plus a Paschen limit photon, etc. There is literally an infinite number of possibilities because there are an infinite number of bound states for the hydrogen atom.

"go into the energy required to keep the proton fixed, or somewhere else?" This doesn't actually make sense. There's nothing holding the proton fixed. In fact, when you examine spectra from stars you see something called Doppler broadening, where the spectrum is spread out by the random motions of the protons and electrons as they move around due to heat. That said, some of the energy can go into changing the kinetic energy of the center of mass of the combined hydrogen. Not much, though, since the photons produced by the collision don't have much momentum so they can't produce much of a kick. A single Lyman limit photon has a momentum of 13.6 eV/c. The change in velocity that represents for a proton is 4.3 meters per second. Even at room temperature, the root-mean-square velocity of a proton is about 1000 m/s. That means that the change in energy for the proton from that kick would be about $50\,\mu \mathrm{eV}$.