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Aloiza [94]
3 years ago
14

When a hydrogen atom makes the transition from the second excited state to the ground state (at -13.6 eV) the energy of the phot

on emitted is:
Chemistry
1 answer:
viktelen [127]3 years ago
3 0

Answer : The energy of the photon emitted is, -12.1 eV

Explanation :

First we have to calculate the 'n^{th}' orbit of hydrogen atom.

Formula used :

E_n=-13.6\times \frac{Z^2}{n^2}ev

where,

E_n = energy of n^{th} orbit

n = number of orbit

Z = atomic number  of hydrogen atom = 1

Energy of n = 1 in an hydrogen atom:

E_1=-13.6\times \frac{1^2}{1^2}eV=-13.6eV

Energy of n = 2 in an hydrogen atom:

E_3=-13.6\times \frac{1^2}{3^2}eV=-1.51eV

Energy change transition from n = 1 to n = 3 occurs.

Let energy change be E.

E=E_-E_3=(-13.6eV)-(-1.51eV)=-12.1eV

The negative sign indicates that energy of the photon emitted.

Thus, the energy of the photon emitted is, -12.1 eV

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Determine the primary structure of an octapeptide from the following data: acid-catalyzed hydrolysis gives 2 arg, leu, lys, met,
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Complete Question

The complete question is shown on the first uploaded image

Answer:

The sequence of these amino acid would be

          <u>Lue</u>  <u>Arg</u>  <u>Lys</u>  <u>Arg</u>  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

Explanation:

The acid-catalyzed hydrolysis from makes us to understand that the  polypeptides contains 8 amino acids

   From the question we are told that the Edman's reagent releases Leu it means that the N-terminal amino acid would be Lue(i.e Leucine)

     Also from the question we are told that the Carboxypeptidase  released Ser this mean the the C-terminal  amino acid would be Ser ( i.e Serine)

   The Amino acids would in the polypeptides would be arranged like this  

         

         <u>Lue</u> __  __  __  __  __ __  <u>Ser</u>

Now from the question we are told that treatment with cranogen bromide form two peptides.

   Now generally  cranogen bromide divides a peptide on  the C- side(i.e the extreme left side ) of  Met(Methionine)(This is an amino acid obtained by hydrolysis of most common peptides )

   So this means that any peptides  containing Met(Methionine) must have Methionine as a C- terminal amino acid(i.e at extreme left) and for peptides that does not contain Met must be C - terminal peptides

From the question we see that it is the second peptide that contain Met and it is a penta peptide(i.e it contains 5 amino acid)  

    Thus the fifth amino acid is  Met  

So the sequence of these amino acid would now be  

          <u>Lue</u> __  __  __  <u>Met</u>  __ __  <u>Ser</u>

From the question we are told that the the Trypsin-catalyzed hydrolysis forms two amino acid and two peptides  

   Now generally Trypsin divides a peptide  on the  C- side(i.e the extreme left) of Arg(Arginine) and Lys (lysine) and any peptide that holds Arg or Lys  must have them as their C- terminal  amino acids

   From the first peptide in the two peptide formed we see that  Arg would be the Seventh amino acid of the octapeptide because commonly the trypsin that sticks to the C-side of Arg would for Ser

and Phe would be the sixth amino acid of the octapeptide

So the sequence of these amino acid would be

       <u>Lue</u> __  __  __  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

Looking  at the first amino acid formed from the Trypsin-catalyzed hydrolysis  we see that Arg would be the fourth amino acid of  octapeptide      as Trypsin divides a peptide  on the  C- side(i.e the extreme left) of Arg(Arginine).

From the second peptide of the  Trypsin-catalyzed hydrolysis  we see that Lys would be the third amino acid of the octapeptide as trypsin divides on the  C- side of  Lys (lysine) and Tyr would be the second amino acid of the octapeptide

So the sequence of these amino acid would be

          <u>Lue</u>  <u>Arg</u>  <u>Lys</u>  <u>Arg</u>  <u>Met</u>  <u>Phe</u> <u>Arg</u>  <u>Ser</u>

           

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