A 100 eV electron collides with a stationary helium ion (He+) in its ground state and excites to a higher level. After the collision the He+ ion emits two photons in succession with wavelength 1085 A∘ and 304 A∘. Find the principal quantum number of the excited state.Given h=6.63×10−34 J.s

A 100 eV electron collides with a stationary helium ion (He+) in its g

1.	A 100 eV electron collides with a stationary helium ion (He+) in its ground state and excites to a higher level. After the collision the He+ ion emits two photons in succession with wavelength 1085 A∘ and 304 A∘. Find the principal quantum number of the excited state.Given h=6.63×10−34 J.s
Answer» 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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A 100 eV electron collides with a stationary helium ion (He+) in its ground state and excites to a higher level. After the collision the He+ ion emits two photons in succession with wavelength 1085 A∘ and 304 A∘. Find the principal quantum number of the excited state.Given h=6.63×10−34 J.s

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