Question

A 100 eV electron collides with a stationary helium ion (He⁺) in its ground state and exits to a higher level. After the collision, He⁺ ions emit two photons in succession with wavelengths 1085 Å and 304 Å. The energy of the electron after the collision will be ______ eV.

Given h = 6.63 × 10^-34 Js.

Options

• 45.20

• 46.24

• 47.70

• 40.23

Answer 1

A 100 eV electron collides with a stationary helium ion (He⁺) in its ground state and exits to a higher level. After the collision, He⁺ ions emit two photons in succession with wavelengths 1085 Å and 304 Å. The energy of the electron after the collision will be 47.70 eV.

Explanation:

The energy of the electron in the n^th state of the He⁺ ion of atomic number Z is given by

E_n = -(13.6) eV `Z^2/n^2` for H⁺ ion Z = 2

Therefore,

`E_n = -((13.6  eV) xx (2)^2)/(n^2)`

= `-54.4/n^2` eV

The energies E₁ and E₂ of the two emitted photons in eV are

`E_1 = 12431/1085` eV = 11.4 eV

and `E_2 = 12431/304` eV  = 40.9 eV

Thus, total energy

E = E₁ + E₂

= 11.4 + 40.9

= 52.3 eV

Let n be the principal quantum number of the excited state.

Now, we have for the transition from n = n to n = 1.

`E = -(54.4) eV(1/1^2 - 1/n^2)`

But E = 52.3 eV

Therefore 52.3 eV = `54.4 eV xx (1/1^2 - 1/n^2)`

or `1 - 1/n^2 = 52.3/54.4 = 0.96`

Which gives n² = 25 or n = 5

The energy of the incident electron = 100 eV (given). The energy supplied to He⁺ ion = 52.3 eV.

Therefore, the energy of the electrons left after the collision = 100 - 52.3 = 47.7 eV.