l T-Mobile LTE 1:35 PM webcampus.fdu.edu 1 (20 pts) Quinine is an antimalaria dr

Question

l T-Mobile LTE 1:35 PM webcampus.fdu.edu 1 (20 pts) Quinine is an antimalaria drug. Under certairn circumstance Quinine can cmit fluorescent light. If a single molecule of quinine emits a single photon having a wavelength of 3475 nm answer the following questions: (i) What is the frequency, in s, of this radiation? (ii)What is the wavenumber, in cm, of this radiation? (in)What is the energy, in J, of a single photon of this radiation? (iv) If an object absorbs a total of 2.85 x10of this radiation, how many photons of this light were absorbed? 2.(30pts) (0) 25pts Which of the following sets of (n.l, m, m) quantum numbers for the H atom are correct. If a given set is incorrect state why this is so (a) 5,2-2,1/2 (b) 6,3,4,-1/2 (c) 3,2,2,-1/2 (d) ,1,12 (e4.4.-2,-1/2 (ii) Spts. In class, the branching diagram for the n-1 and n- 2 cases were developed. Write the branching diagram for the n-3 case. Clearly show the values of the various quantum numbers. 3.(20pes) Calculate the wavelength, in nm, of the light emitted when a B" ion makes a transition from the n=5 to the n= 3 state. Hint: The atomic number of B4" is Z-5, and it has one electron. 4.(12pts) dentify the group of elements that correspond to each of the following generalized electron configurations. No partial credit. You should chock your work using table 7.3 ofet [Noble gas] ns(n-2)f ics [Noble gas] ns(n-1 n(Noble gas] nsnp ,. GENERAL CHEMISTRY I HOMEWORK ASSIGNMENT PRIVATE (continued) (18ps) (a) 6pts. What is the ground state electron configuration and orbital diagram of Ta(73)? Is this paramagnetic or diamagnetic? (b) 4pts. The following is an excited state configuration of a neutral atom: Identify the element and write its ground state electronic configuration. Is this excited state diamagnetic or paramagnetic? Explain to 8pts, Write the ground state electron configuration of each of the following ions: (OMn (25) (ii) Rh"(45) ) Pr (78) (iv) As' (33)

Explanation / Answer

Q1

a)

relate wavlenegth to frequency via

WL = c/v

v = c/WL

where c = speed of light = 3*!0^8 m/s

v = frequency in seconds, WL = wavleneght in meter

v = (3*10^8)/(347.5*10^-9)

v =8.633*10^14 Hz or s^-1

ii)

wavenumbr is simply -->

reciprocal of Wavleenght, into cm

Wl = (347.5 nm ) = 347.5*10^-7 cm

wave no = 1/(347.5*10^-7) = 28776.97 cm^-1

iii)

Energy of 1 photon is given by

E = hv

h = Planck Constant = 6.626*10^-34 J s

E = (6.626*10^-34)(8.633*10^14) = 5.7202*10^-19 J/photon

iv)

if E absorbed = 2.85*10^-9 J

find photons

N = Etotal / Ephoton = ( 2.85*10^-9) / ( 5.7202*10^-19 ) =

4.98*10^9 photons

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