Question

1. It turns out that while Bohr’s model and equation for the calculation of emission energies does not work for atoms larger than hydrogen, it does work for any atom with only a single electron, so long as we include a factor of Z2 in the Rydberg expression, where Z is the charge on the nucleus. Set up an excel spreadsheet to calculate the difference in energy, wavelength, and frequency corresponding to the following transitions:

a. Hydrogen n = 1 to 2

b. Hydrogen n = 2 to 3

c. Hydrogen n =3 to 4

d. Hydrogen n = 9 to 10

e. Hydrogen n = 9 to 5

f. Rb+36 n = 2 to 3

g. Xe+53 n = 2 to 3

h. Pa+90 n = 2 to 3

i. Explain any trends you observe. (submit your excel on BB).

I will be very thankful for any and all help!!!

Answer 1

1/=R(1/n1^2-1/n2^2)

R=rydberg constant=1.097*10^7 m-1=1.097*10^5 cm-1

n1,n2 are principal quantum numbers of orbitals occupied before and after transition of an electron

=wavelength of light

For hydrogen-like atom. rydberg formula- 1/=RZ^2(1/n1^2-1/n2^2) [Z=atomic number]

frequency==c/ c=speed of light=3.00*10^10 cm/s

As light of frequency is emitted on transition, this corresponds to the energy of transition of electron from one orbital to other,E=h ,h=planck's constant=6.63*10^-34 m2/Kg s

Calculations

1)n=1 to 2

1/=R(1/n1^2-1/n2^2)

1/=1.097*10^5 cm-1 (1/1-1/2^2)

or,1/=1.097*10^5 cm-1 (1/1-1/4)

or,1/=1.097*10^5 cm-1 *0.76=0.823*10^5

or,=1/0.823*10^5=1.215*10^-5 cm-1

frequency==c/=(3.00*10^10 cm/s)/(1.215*10^-5 cm-1)=2.47*10^15 s-1

Energy difference=h ,h=planck's constant=6.63*10^-34 m2/Kg s

or,E=(6.63*10^-34 m2/Kg s) *(2.47*10^15 s-1)=16.38*10^-19 J

Similarly. 2)

n=2 to 3

1/=R Z^2(1/n1^2-1/n2^2)

1/=1.097*10^5 cm-1 * (37)^2* (1/2^2-1/3^2)

or,1/=1.501*10^8 cm-1 (1/4-1/9)

or,1/=1.501*10^8 cm-1 *0.139=0.209*10^8 cm-1

or,=1/0.209*10^8=4.785*10^-8 cm-1

frequency==c/=(3.00*10^10 cm/s)/(4.785*10^-8 cm-1)=0.627*10^18 s-1

Energy difference=h ,h=planck's constant=6.63*10^-34 m2/Kg s

or,E=(6.63*10^-34 m2/Kg s) *(0.627*10^18 s-1)=4.157*10^-16 J

f) Z=37 (Rb) atomic no

n=2 to 3

1/=R(1/n1^2-1/n2^2)

1/=1.097*10^5 cm-1 (1/2^2-1/3^2)

or,1/=1.097*10^5 cm-1 (1/4-1/9)

or,1/=1.097*10^5 cm-1 *0.139=0.152*10^5

or,=1/0.152*10^5=6.579*10^-5 cm-1

frequency==c/=(3.00*10^10 cm/s)/(6.579*10^-5 cm-1)=0.456*10^15 s-1

Energy difference=h ,h=planck's constant=6.63*10^-34 m2/Kg s

or,E=(6.63*10^-34 m2/Kg s) *(0.456*10^15 s-1)=3.023*10^-19 J

trend observed by comparing the values of transition from n=1 to 2 and n=2 to 3 for hydrogen atom is that energy difference between orbitals 1 and 2 is less compared to that between 2 and 3,or energy difference increases for higher orbitals