For parts A and B consider a doubly ionized lithium atom (Li^2+, Z=3) a) Find th

Question

For parts A and B consider a doubly ionized lithium atom (Li^2+, Z=3)

a) Find the radius and energy of the electron in the n= 3 state of the atom

b) This electron now jumps to the n = 2 state. How does the wavelength of the emitted photon in this transition compare to a photon emitted by the same transition in hydrogen atom? ( Calculate ratio)

c)The electron in a hydrogen atom with an energy of -0.544 eV is in a subshell with 18 states. Find the corresponding principal quantum number, n.

d) An electron drops from the L shell to the K shell and gives off an Xray with a wavelenth of .0205 nm What is the atomic number of this atom?


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Explanation / Answer

a)   The radius of the electron in the n =3 state of the atom is                     rn = (n^2/Z)ao                     r3 =[3^2/3](0.0529 nm)                          = 0.1587 nm The energy of the electron in the n =3 state of the atom is           En = - Z^2 (13.6 eV)/n^2                = - (3)^2[13.6 eV]/9                 = -13.6 eV b) The wavelength of photon when electron jumps from n=3 state to n=2 state is               Li = (Z^3 R[1/2^2 -1/3^2])-1                  Here Z = 3           R = 1.097 x 10^7 /m                                Li = 72.9 nm For hydrogen atom         H = 656.112 nm    So Li / H =(72.9 nm)/( 656.112 nm)                        = 0.11
c) The energy of the electron in the nth state is                En =-(13.6 eV)/n^2 Here En = -0.544 eV                n = [(-13.6 eV)/( -0.544 eV )]                   = 5
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