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, $H e^{+}$ ion emits two photons in succession with wavelengths $1085 ˚ A$ and $304 ˚ A$ . Find the principal quantum number of the excited state. Also, calculate the energy of the electron after the collision. Given $h = 6.63 \times 10^{- 34} J s$ .

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Solution

Energy of $H e^{+}$ in ground state $= \frac{- 13.6}{1^{2}} \times 4 = - 54.4$
Total energy before collision $= 100 - 54.4 = 45.6 e V$
From conservation of energy:
$45.6 = E + \frac{h c}{1085} + \frac{h c}{304} E = 45.6 - 11.42 - 40.78 E = 45.6 - 52.2 E = - 6.6 \frac{- 13.6}{n^{2}} \times 4 = - 6.6 n^{2} = 8.2 n \approx 2$
Principal quantum number = 2

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Q. 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

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)