Question

Calculate the energy in calories required to produce, from neutral He atoms,1 mole of He+ ions and He++ ions using Bohr's equations.

Answer 1

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Answer 2

Step-by-step explanation:

Energy necessary to remove an electron from outer orbital of a neutral gaseous atom or molecule is known as ionization energy.

Bohr's equation to calculate energy is as follows.

`\Delta E = R_(H)[(1)/(n^(2)_(i)) - (1)/(n^(2)_(f))]`

where,
`R_(H)` = Reydberg's constant =
`1.09 * 10^(7)` per meter

In the given case,
`n_(1)` = 1 and
`n_(2)` =
`\infty`

Therefore, calculate the energy as follows.

`\Delta E = R_(H)[(1)/(n^(2)_(i)) - (1)/(n^(2)_(f))]`

=
`1.09 * 10^(7)[(1)/((1)^(2)) - (1)/(\infty)}]` J

=
`1.09 * 10^(7)[1 - 0}]` J

=
`1.09 * 10^(7)]` J

As atomic configuration of helium is
`1s^(2)`. Thus, energy required to produce
`He^(+)` and
`He^(2+)` will be the same.

Also, 1 joule = 0.239 calories. Hence, convert calculated energy into joules as follows.

`1.09 * 0.239 * 10^(7)` J

=
`0.26051 * 10^(7)` cal

=
`26.051 * 10^(5)` cal

Therefore, we can conclude that
`26.051 * 10^(5)` cal energy is required to produce, from neutral He atoms,1 mole of
`He^(+)` ions and
`He^(2+)` ions using Bohr's equations.