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(a) Suppose you put both electrons in a helium atom into the n=2state;

what would the energy of the emitted electron be?

(b) Describe (quantitatively) the spectrum of the helium ion,He+.

Short Answer

Expert verified

(a)The energy of the emitted electron is 2E1-4E1=-2E1=27.2eV

(b)Helium has one electron and it鈥檚 a hydrogenise ion with z=2 so the spectrum

is1/=4R(1/nf2-1/ni2)

Step by step solution

01

(a) The energy of the emitted electron

The energy of each electron isE=Z2E1/n2=4E1/4=E1=E1=-13.6eV,

so the total initial energy is2-13.6eV=-27.2eV.

One electron drops to the ground state Z2E1/1=4E1, so the other is left

with 2E1-4E1=-2E1=27.2eV.

02

(b) The spectrum of the helium ion

(b) He+has one electron; it鈥檚 a hydrogenise ion with Z = 2, so the spectrum

is 1/=4R1/nf2-1/ni2, where R is the hydrogen Rydberg constant, and ni,nfare the

initial and final quantum numbers (1, 2, 3, . . . ).

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