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Ask iit jee aieee pet cbse icse state board community Community Discussion Question: help !! atomic structure
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shadowfurion (0)

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1) find
    a)total number of
    b)total mass of protons of 34 mg of Nh3   AT STP . WILL ANSWER CHANGE IF  
        
TEMPERATURE AND  PRESSURE IS CHANGED  ?

2) the electron in hydrogen atom is given by  E =  (- 21.7 x 10 -12      divided by  n  2  ) ergs. Calculate the energy required to remove an electron completely from the n = 2 orbit . what is the longest wave length ( in cm?s ) of light that can be used to cause this transition.
 


    
catch_arnnie (521)

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for question 1),
i don't think the ans. will change with change in temperature & pressure coz' we are talking of proton which is in the nucleus of atom, so even with change of temperature & pressure, the proton will stay there.
 
now, 1 molecule of NH3 has 1 proton each of 3 Hydrogen & 14 proton of 1 Nitrogen, so, total protons  = 17 per molecule
 
1 mole NH3 is 28g
 
=> 28g of NH3 will have 6.023 * 1023 molecules
=> 34mg of NH3 will have 6.023 * 1020 * 34 / 28 molecules
 
so, total no. of protons(say p) = 6.023 * 1020 * 34 * 17 /28  = 124.3 * 1020 protons
 
total mass of protons = p * (mass of 1 proton)
 
i think you can do the last part on your own...
 
plz correct me if i'm wrong anywhere....
 
 
 
 

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akku (1099)

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energy req 2 remov the en completely
E=E2-E= - 21.7*10^-12(1/4-1/)= -5.4*10^-3 ergs =energy by which the en is bound =-energy req 2 remov it=5.4*10^-3 ergs
E=hc/# =hc(1/22 -1/32)
#=wavelength(longest)=hc/E=6.63*10^-34*3*10^8*5/36/5.4*10^-3*10^-7 (converting ergs 2 joules)

sry I've corrected my post had overlooked the word 'longest'
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sachin_gupta1991 (69)

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g1)
 
1 molecule of NH3 has 1 proton each of 3 Hydrogen & 14 proton of 1 Nitrogen, so, total protons  = 17 per molecule
 
1 mole NH3 is 17g.
We have 0.002 moles , i.e. ,therefore we have 6.022 x 1023 x 0.002=12.044 x 1020 molecules.
And  no. of protons=12.044 x 1020 x 17
Mass of protons= 12.044 x 1020 x 17 x 1.66 x 10-24 g  {Becoz mass of a proton=1.66 x 10-24 
 
which gives 0.034 g.
Plz rate me if correct.                    
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somyekathait (202)

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fo question 2)
part a)
E=(21.7 X 10^-12) / 4
=5.42 X 10^ -12
part b)
5.42 X 10^ -12 = hc/lembda
where h = 6.6 X 10 ^ -27 and c =3 X 10^8
on substitutung values in this we get lembda =3.65 X 10 ^ -5 cms

I WAS BORN INTELLIGENT!!!!!!!
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