1. A 386 g sample of water absorbs infrared radiation at 1.06 x 10^4 nm from a c

Question

1. A 386 g sample of water absorbs infrared radiation at 1.06 x 10^4 nm from a carbon dioxide laser. Suppose 45.0% of the absorbed radiation is converted to heat. Calculate the number of photons at this wavelength required to raise the temperature of the water from 25.0 degrees C to 40.0 degrees C

2. For the principal quantum number n=4, write out all possible values of l and ml and ms
    b) Put all of the orbitals yhou identified in part a in order of increasing energy according to the Aufbau process. Use the designations 1s, 2s etc for this question. You may include orbitals for wich n>4 in order to complete the series.
    c) Determine an ion on the periodic table that has a +3 charge when there are 7 electrons in the subshell n=4, 1=2. Explain your answer.

Explanation / Answer

Q. 1.

Step 1 :

Calculation of heat energy required to raise the temperature of 386 g water by (40.0 -25.0 ) degree.

Formula:

q = m x C x delta T

q is heat energy in J, m is mass in g, C is specific heat of water, Delta T is change in T.

Lets plug all value to get q.

q = 386 g x 4.184 J / g 0C x 15 0C

=24225.36 J

So 24225.36 J is the energy absorbed from the photon.

Calculation of photon in this much energy

Number of photons = Total energy / energy of single photon

Step 2

Calculation of energy of photon

E = h c /l

Here h is planks constant = 6.626 E-34 J s

c is speed of light =3.0 E8 m/s

l = wavelength in m.

Solution we are given wavelength in nm. Lets convert it to m

Conversion of wavelength in m

l= 1.06E4 nm x 1 E-9 m / 1 nm

=1.06E-5 m

Lets plug and calculate Energy

E = ((6.626*10^-34)*3.0*10^8/1.06*10^(-5))

E = 1.87 E-20 J / photon

Calculation of number of photon

Number of photon = 24225.36 J / 1.87 E-20 J / photon

=1.29E24 photons

But actually these are the only 45 % of original so we have to calculated original number of photons .

Original number of photons needed to raise the temperature of water

= 100 x (1.29E24 / 75 ) photon

= 1.72E24 photons

So the total number of photon required = 1.72 E24

Q. 2

Lets show possible values of l and ml and ms

n = 4

l = 0   (orbital is   4s) , ml = 0 , ms = +/- ½

l = 1 (Orbital is p , 4p ), ml = (-1, 0, +1), ms =+/- ½

l = 2 (oribital is 4d) , ml = (-2 -1, 0, +1 +2 ), ms =+/- ½

l = 3 (orbital is 4f) ml = (-3 -2 -1, 0, +1, + 2, + 3 ), ms =+/- ½

So the orbital energy (increasing order)

4s 3d   4p   5s    4d    4f

Q. c

Ion has +3 charge

7 electrons in shell n = 4 ,

There are 18 electrons in outer shell of n = 4

But there are only 7 electrons are there. We also add 3 electrons since ion has 3+ charge.

So total electrons = 10

Now Atomic number of this ion = Atomic number of Kr + 10 electrons          ( Kr is last for n = 3)

So the cation has atomic number = 46

And that is Pd

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