A $100 eV$ electron collides with a stationary helium ion $(H e^{+})$ in its ground state and excites to a higher level. After the collision the $H e^{+}$ ion emits two photons in succession with wavelength $1085 A^{\circ}$ and $304 A^{\circ}$ . Find the principal quantum number of the excited state.
Given $h = 6.63 \times 10^{- 34} J.s$

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Solution

The energy of the electron in the $n^{t h}$ state of $H e^{+}$ ion of atomic number $Z$ is given by:
$E_{n} = - (13.6 e V) \frac{Z^{2}}{n^{2}}$
For $H e^{+}$ ion , $Z = 2$ therefore;
$E_{n} = - (13.6 e V) \times \frac{4}{n^{2}}$
$\Rightarrow E_{n} = - \frac{54.4}{n^{2}} e V . . . . (1)$
The energies $E_{1}$ and $E_{2}$ of the two emitted photons in $e V$ are,
$E_{1} = \frac{12431}{1085} e V = 11.4 e V$
$E_{2} = \frac{12431}{304} e V = 40.9 e V$
Thus total energy is,
$E = E_{1} + E_{2} = 11.4 + 40.9 = 52.3 e V$
Let $n$ be the principal quantum number of the excited state. Using the equation $(1)$ we have for the transition,
$\Rightarrow E = - (54.4 e V) (\frac{1}{1^{2}} - \frac{1}{n^{2}})$
substituting $E = 52.3 e V$ we get,
$52.3 e V = 54.4 e V (1 - \frac{1}{n^{2}})$
$\Rightarrow (1 - \frac{1}{n^{2}}) = \frac{52.3}{54.4} = 0.96$
$\Rightarrow n^{2} = 25$
$Or, n = 5$

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Q. A

100 e V

electron collides with a stationary helium ion

(H e^{+})

in its ground state and excites to a higher level. After the collision,

H e^{+}

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1085 ˚ A

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. Find the principal quantum number of the excited state. Also, calculate the energy of the electron after the collision. Given

h = 6.63 \times 10^{- 34} J s

Q. Find the quantum number

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During the emission spectrum of electron in $H e^{+}$ ion, the electron on transition from same higher energy state drops to the ground state and first excited state emits two photon of $λ_{1}$ =304 $A^{\circ}$ and $λ_{2}$ =1085 $A^{\circ}$ respectively.The total number of photons emitted during the transition of electrons from highest excited to all lower energy level(Possible)