Showing posts with label Energy. Show all posts
Showing posts with label Energy. Show all posts

Numericals on Radius, Energy, Frequency, Wavelength and Wave Number


Numericals on Radius, Energy, Frequency, wavelength and wave number from Textbook


Q1.  Find the radius of 4th orbit of electron in Hydrogen atom (Book question; page # 24, Example # 2.1)

Solution

Z   = 1 (for H atom)

‘n’ = 4 (4th orbit)

‘ao’= 0.529 Å

 


Q2. Calculate the energy of an electron in L-shell of hydrogen atom. The value of K 2.18 x 10−18 J/atom 

(Book question; page # 25, Example # 2.2)


Solution


‘n’ = 2 (L-shell corresponds to second energy level)

‘K’= 2.18 x 10−18 J/atom








Q3. Using Bohr model, determine the energy in joule of a photon produced when an electron in hydrogen atom jumps from an orbit n=5 to n=2. 

(Book question; page # 26, Self-Assessment)


Solution

‘n2’ = 5 

‘n1’ = 2






Q4. Calculate the wave numbers of photons when electron of a hydrogen atom jumps form 4th orbit to 2nd orbit. value of RH = 1.09678 x 107 m−1.  

(Book question; page # 27, Example 2.3)


Solution

‘n2’ = 4 

‘n1’ = 2

Z    = 1

RH = 1.09678 x 107 m−1






Q5. What is the wave number of a photon produced when an electron falls from n=5 level to n=3 level in hydrogen atom. value of RH = 1.09678 x 107 m−1.  

(Book question; page # 28, Self-Assessment)


Solution

‘n2’ = 5 

‘n1’ = 3

Z     = 1

RH = 1.09678 x 107 m−1














HOT FAVOURITE QUESTION


*Q6*. A photon of wave number 23 x 105 m−1 is emitted when electron undergoes a transition from a higher orbits to n=2. Determine the orbit form which electron falls and also the spectral line appears in this transition of electron (The value of Rydberg constant is 1.09678 x 107 m−1).


(Book question; page # 49, Assignment)


Solution

RH = 1.09678 x 107 m−1

Z   = 1











Numericals on Radius, Energy,Frequency, Wavelength and Wave Number from External Source

 



Q1.    If the radius of first Bohr orbit is ‘x’ then calculate the radius of the third orbit of H atom. 
Solution




For given part 
‘n’ = 1 
ao = r1= 0.529 Å= ‘x’ 
r3 = ? 
 
Radius (r3) = (x) x  32/1 Å      Þ = (x) x  9/1 Å      Þ r3 = 9x Å      
 

The radius of the third orbit of H atom is 9 times than that of radius of first orbit. 


Q2.  If the radius of first Bohr orbit is ‘x’ Å then calculate the radius of the 2nd orbit of hydrogen. 
Solution

Z = 1 (for H atom)

‘n’ = 2 (2nd orbit)

‘ao’= ‘x’ Å







The radius of the second orbit of H atom is 4 times than that of radius of first orbit. 



Q3.Calculate the ratio of radius of second and third orbit of hydrogen atom. (Bohr’s radius =             0.529 Å)

Solution

Z = 1 (for H atom)

‘n’ = 2 (2nd orbit)

‘n’ = 3 (3rd orbit)

‘ao’= 0.529 Å





 

 

Q4. Calculate the ratio of radius of third and fourth orbit of hydrogen atom. (Bohr’s radius = 0.0529 nm)

Solution

Z = 1 (for H atom)

‘n’ = 3 (3rd orbit)

‘n’ = 4 (4th orbit)

‘ao’= 0.0529 nm



 



Q5.Calculate the ratio of the radius of Li2+ ion in 3rd energy level to that of He+ ion 2nd energy level.

Solution 











Q6.What is the wavelength and wave number of radiation that is emitted when a hydrogen atom undergoes a transition from orbit 3 to orbit 1.

Solution

 ‘n2’ = 3 

‘n1’ = 1

Z              = 1

RH  = 1.09678 x 107 m−1

 

Calculation of Wave Number











Calculation of Wave Length 





Q7. Calculate the wave number and wavelength of Balmer series of hydrogen spectrum in which electron jumps from orbit 3 to orbit 2.

 

Solution

Calculation of Wave Number

n1  = 2

n2  = 3

RH = 1.0968553 x 107 m-1

Z   = 1













Q8.  Calculate the energy of the electron in the ground state and first excited state of the hydrogen atom

Solution

Z for H = 1

n for ground state = 1

n for first excited state = ground state +1 = 1+1=2 (corresponds second energy level)

‘K’= 2.18 x 10−18 J/atom or 13.6 eV










Q9. The energy of electron in the excited state of hydrogen atom is -0.85 eV.  calculate the value of ‘n’ for in the excited state.

Solution 








Assignment


Q1.  Calculate the energy of 1st, 2nd and 3rd orbit of hydrogen atom.

                (Answers; -13.6 eV, -3.41 eV, -1.51 eV)

Q2.  Calculate the radius of 1st, 2nd and 3rd orbit of hydrogen atom.

                (Answers; 0.529°A, 2.226°A, 4.761°A)

Q3. Calculate the angular momentum of 1st and 2nd orbit of hydrogen atom.

                (1.054 x 10−34 Js, 2.108 x 10−34 J.s)

 

Q4. Calculate the wave number of the line in Lyman series when an electron jumps from orbit 3 to orbit 1.

                (Answer; 97613.4 cm−1)

Q5. Calculate the wave number of spectral line of hydrogen gas when an electron jumps from n = 4 to n = 2.

                (Answer; 20654.6 cm−1)















Derivation of Radius, energy, energy difference, frequency, wave no, and wavelength of the nth Bohr’s Orbit for Hydrogen-Like Atoms

 

Derivation of Radius of the nth Bohr’s Orbit for Hydrogen-Like Atoms


Assumptions for simple atom

To derive an expression for radius, consider a hydrogen atom (or hydrogen-like atoms such as He+, Li2+, Be3+, B4+, C5+) with atomic number equal to z consisting of a single electron with charge –e and mass m revolving around the nucleus of charge +Ze (+e is charge of proton) with a tangential velocity v in the orbit whose radius is r.

Now revolving electron is being acted upon simultaneously by the following two types of forces;




(i) Electrostatic force of Attraction / Centripetal Force

According to Coulomb’s law, the electrostatic force of attraction (Fe) between the nucleus of charge ‘+Ze’ and electron of charge ‘–e’ separated by a distance ‘r’ is given by:






Where ‘K’ is proportionality constant. It is equal to 1/4πεor2

Hence attractive force between nucleus and electron can be written as






(ii) Centrifugal Force

This Coulombic force of (Fc) supplies the centrifugal force to keep the electron in an orbit and is given by:





Equating Fe and Fc

To keep the electron in the same orbit, these two opposite forces must be equal to each other i.e.




Determination of v2 of electron by using Bohr’s Postulate

According to Bohr’s postulate, angular momentum of electron revolving around the nucleus is an integral multiple of h/2p.





Calculation of Radius (r)

Substituting the value v2 from equation (ii) in equation (i)






Where

h    = Planck’s constant = 6.625 x 10−34 J.s

me = Mass of electron = 9.11 x 10−31 kg

e    = Charge of electron = 1.602 x 10−19 C

𝛆= Vacuum permittivity constant = 8.84 x 10−12 C2/J.m

 

Calculation of nth Bohr’s orbit (rn)

Assembling all constants in equation (iii), we get



Where a is known as Bohr’s constant or Bohr radius and its value is 0.529 x 10−8 cm or 0.529 Å or 0.0529nm or 52.9 pm. This is the radius of the first orbit of H. This equation is used for the determination of nth orbit of hydrogen atom and hydrogen like ions like He+, Li2+ etc. 


The above equation shows that radius of orbit is directly proportional to the square of the principal quantum numbers (r α n2 i.e. 1, 2, 3, ……..) and inversely proportional to atomic number. As the value of n increases, the radius of the orbit will increase.

 






 

Derivation of Energy of the nth Bohr’s Orbit


Basic of Derivation

The total energy of an electron revolving in any orbit around the nucleus is the sum of kinetic energy and potential energy given by,

Etotal = K.E + PE ……………….(i)

 

Calculation of K.E

The K.E. of electron with mass m revolving around the nucleus with velocity v is given by the following expression;



Now the centrifugal and centripetal forces upon the revolving electron are given as:



At uniform circular equilibrium motion, these two opposite forces must be equal to each other i.e.






Calculation of P.E.

P.E is the work done in bringing the electron from infinity to a point at a distance r from nucleus and can be calculate as

P.E =work done = −force x displacement = −Fe x r





Here negative sign indicates that P.E decreases when electron is brought form infinity to a point at a distance r. Here negative sign indicates a net attractive interaction, giving algebraically lower energy at shorter distance.


Calculation of Total Energy

Etotal = K.E + PE















Here K is a factor assembled by various constant present in energy equation. Its value is 2.18 x 10−18 J/atom or 1312.8 kJ/mol 


E is always negative. Negative sign shows that the electron is bound to the atom and energy must be spent in order to remove it from the orbit. 

All energy states are bound states as the negative sign indicates. When n = 1; this corresponds to electron at the closest possible distance from the nucleus and at its lowest energy and is called ground state energy. All energy states with value of n higher than 1 are termed as excited states. When n = α then E = 0; which means that the system is unbound and the electron is free. 


It should be noted that the energy is increasing as the n (orbits) increasing; however the difference of energy between two orbits is decreasing. 



Conclusion

If total energy = − x

Then

KE = + x

PE = − 2x 



 

2.12 Expression for ∆E of electronic transition between orbits

 

Calculation of ∆E

Let E1 be the energy of n1 orbit and E2 is for n2 orbit. To calculate the energy emitted by atom in the form of radiation when an electron jumps from a higher energy state n2 to lower energy orbit n1; let us make use of the postulate of Bohr’s model; according to which, the emitted energy is written as

 Emitted energy = ∆E = E2 – E1



 

But,









This equation is used for determining the emission or absorption of energy when electron jumps from one orbit to another 



2.13 Expression for Frequency of electronic transition between orbits

 

Calculation of ∆E

Let E1 be the energy of n1 orbit and E2 is for n2 orbit. To calculate the energy emitted by atom in the form of radiation when an electron jumps from a higher energy state n2 to lower energy orbit n1; let us make use of the postulate of Bohr’s model; according to which, the emitted energy is written as


Emitted energy = ∆E = E2 – E1

But,


To calculate the frequency (u) of emitted radiations (or photons there in); let us make use of the Bohr’s postulate; according to which:


hu = ∆E = E2 – E1

But 






 


To calculate the frequency (u) of emitted radiations (or photons there in); let us make use of the Bohr’s postulate; according to which:

hu = ∆E = E2 – E1

 

But 











 

2.14 Expression for Wave Number and Wave Length of Radiation

 
























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