Chemistry color{red} ✎ PROBLEM SOLVING TECHNIQUES
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### Topics covered

• Problems based on nuclear structure
• Problems based on Planck's quantum theory and photoelectric effect
• For Balmer series [ Rydberg formula ]
• Problems based on Bohr's model
• Problems based on De- broglie
• Problems based on Heisenberg's Uncertainty principle
• Problems based on electronic configuration

### PST-1 Problems based on nuclear structure

Q 3166878775

Complete the following table :

Particle Mass Number Atomic Number Protons Neutrons Electrons
Nitrogen atom - - - 7 7
Calcium ion - 20 - 20 -
Oxygen atom 16 8 - - -
Bromide - - - 45 36

Solution:

For nitrogen atom :

No. of electrons = 7 ( given )

No. of neutrons = 7 ( given)

therefore No. of protons = Z = 7 ( because atom is electrically neutral).

Atomic number = Z = 7

Mass no. (A) = NO. of protons+ NO. of neutrons

 = 7+7 = 14

For Calcium ion

No. of neutrons = 20 ( given)

Atomic Number (Z) = 20 ( given)

therefore No. of protons  = Z = 20

No. of electrons in calcium atom = Z = 20

But in the formation of calcium ion , two electrons are lost from the extranuclear part according to the equation Ca → Ca^(2+) + 2e^(-)
but the composition of the nucleus remains unchanged.

therefore No. of electrons in calcium ion = 20-2 = 18

Mass number(A) = No. of protons + No. of neutrons

 = 20+20= 40

For oxygen atom :

Mass number (A) = No. of protons +No. of neutrons = 16 (given)

Atomic No. (Z) = 8.(given)

No. of protons = Z = 8 (given)

No. of electrons = Z = 8

No. of neutrons = A-Z = 16-8 = 8

For bromide ion : In the formation of bromide ion one electron is added to the extra nuclear part of the bromine atom according to the equation Br + e^(-) → Br^(-) but the composition of the nucleus remains unchanged .

Since the no. of electrons in bromide ion = 36 (given)

No. of electrons in bromine atom  = 36-1 = 35

Further , since in the above reaction , nucleus of the bromine atom remains unchanged.

therefore No. of protons in the bromine ion = No. of protons in the bromine atom = 35.

Putting the values of the various vacant place as calculated above , in the table we have

Particle Mass Number (A) Atomic Number (Z) No. of Protons(Z) No. of Neutrons(A-Z) No. of Electrons(Z)
Nitrogen atom 7+7 = 14 7 7 7
Calcium ion 20+20= 40 20 20 20 20-2 = 18
Oxygen atom 16 8 8 16-8 = 8 8
Bromide ion 45+35 = 80 35 36-1 = 35 45 36

### PST-2 Problems based on Planck's quantum theory and photoelectric effect

Q 3116180970

Calculate the kinetic energy of the electron ejected when yellow light of frequency 5.2xx10^(14) sec^(-1) falls on the surface of potassium metal . Threshold frequency of potassium is 5xx10^(14) sec^(-1)

Solution:

K.E of the ejected electron is given by

1/2 mv^2 = hv - hv_0 = h(v - v_0)

 = 6.625xx10^(-34) xx (5.2xx10^(14) -5.0xx10^(14))

 = 6.625xx10^(-34) xx 0.2 xx 10^(14) Joules

= 1.325 xx 10^(-20) Joules

### PST-3 For Balmer series [ Rydberg formula ]

Q 3146191073

Calculate the frequency and the wavelength of the radiation in nanometers emitted when an electron in the hydrogen atom jumps from third orbit to the ground state. In which region of the electromagnetic spectrum will this line lie ? ( Rydberg constant = 109 , 677 cm text()^(-1) )

Solution:

According to Rydberg formula

barnu = R ( 1/n_1^2 - 1/n_2^2)

Here R = 109 , 677 cm^(-1)

n_2 = 3

n_1 = 1 ( for ground state)

therefore barnu = 109, 677(1/1^2 - 1/3^2) cm^(-1)

 = 109 , 677 xx 8/9 cm^(-1) = 97490.7 cm^(-1)

lamda = 1/(bar nu) = 1/(97490.7 cm^(-1))

 = 103xx10^(-7) cm

 = 103xx10^(-9) m = 103 nm

nu = c/lamda = (3xx10^8 m s^(-1))/(103xx10^(-9)m)

 = 2.91xx10^(15) s^(-1)

The wavelength as calculated above , lies in this ultraviolet region . Otherwise too , as the jump is on the 1st orbit , the line will belong to Lyman series and hence lie in the ultraviolet region
Q 3116191079

The wavelength of the first line in the Balmer series is 656 nm. Calculate the wavelength of the second line and the limiting line in Barmer series .

Solution:

According to Rydberg formula

barnu = 1/lamda = R (1/n_1^2 - 1/n_2^2)

For the balmer series n_1 = 2 and for the 1st line .

n_2 = 3

therefore 1/656 = R (1/2^2 -1/3^2) = R xx (1/4-1/9)

= R xx 5/36 = (5R)/(36) .................(i)

For the second line , n_1 = 2 , n_2 = 4

therefore 1/lamda = R (1/2^2-1/4^2) = R (1/4-1/16)

 = R xx 3/16 = (3R)/(16) ...........(ii)

Dividing (i) by (ii) we get

lamda/(656) = 5/36xx16/3

or lamda = 485.9 nm

For the limiting line n_1 = 2

n_2 = oo

therefore 1/lamda = R (1/2^2-1/oo^2) = R/4 ......(iii)

Dividing (i) by (iii) we get

lamda/(656) = 5/36 xx 4

or lamda = 364.4 nm

Alternatively first calculate R from (i) and sub stitute in (ii) and (iii)

### PST-4 Problems based on Bohr's model

Q 3146191973

Calculate the wavelength of the radiation emitted when an electron in a hydrogen atom undergoes a transition from 4th energy level to the 2nd energy level. In which part of the electromagnetic spectrum does this line lie ?

Solution:

E_n = - (21.8xx10^(-19))/n^2 J atom text()^(-1)

Energy emitted when the electron jumps from  n= 4 to n = 2 will be given by

DeltaE = E_4 - E_2 = 21.8 xx 10^(-19) (1/2^2-1/4^2)

 = 21.8xx10^(-19) xx3/16 = 4.0875xx10^(-19) J

The wavelength corresponding to this energy can be calculate using the expression

E = hnu = h . c/lamda \ \ \ \ ( because c = nu lamda )

so that lamda = ( hc)/E

= ((6.626xx10^(-34) Js) (3xx10^8 m s^(-1)))/((4.0875xx10^(-19) J))

 = 4.863xx10^(-7) m

= 4863 overset(o)A

or 486.3 nm

It lies in the visible region

### PST-5 Problems based on De- broglie

Q 3166491375

Two particles A and B are in motion. If the wavelength associated with particles A is 5xx10^(-8) m calculate the wavelength associated with particle B if its momentum is half of A.

Solution:

By de Broglie equation

lamda_A = h/(P_A) and lamda_B = h/(p_B)

therefore = lamda_A/lamda_B = p_B/P_A

But p_B = 1/2 p_A ( given)

lamda_A/lamda_B = (1/2 p_A)/(p_A) = 1/2

or lamda_B = 2 xx lamda_A = 2xx5xx10^(-8) m

= 10^(-7) m
Q 3106491378

The kinetic energy of a sub - atomic particle is 5.85xx10^(-25) J. Calculate the frequency of the particle wave. ( Plank's constant , h = 6.626xx10^(-34) kg m^2 s^(-1))

Solution:

K.E = 1/2 mv^2 = 5.85xx10^(-25) J

By de- Broglie equation lamda = h/(mv)

But lamda = v/nu

therefore v/nu = h/(mv)

or nu = (mv^2)/h = ( 2xx5.85xx10^(- 25) J)/(6.626xx10^(-34) J s)

 = 1.77 xx 10^9 s^(-1)

### PST-6 Problems based onHeisenberg's Uncertainty principle

Q 3126591471

Calculate the uncertainty in the velocity of a wagon of mass 3000kg whose position is known to an accuracy of pm 10 p m (Plank's constant = 6.63xx10^(-34) Js)

Solution:

Here we are given m = 3000kg

Deltax = 10 p m = 10xx10^(-12) m = 10^(-11) m

therefore By uncertainty principle

Deltav = h/(4pi xx m xx Deltax)

= (6.63xx10^(-34) kg m^2 s^(-1))/(4xx22/7xx3000 kg xx10^(-11) m)

= 1.76xx10^(-27) m s^(-1)

### PST-7 Problems based on electonic configuration

Q 3156691574

A neutral atom of elements has 2K , 8L and 5M electrons . Field out the following from the data : (a) Atomic No. (b) Total No. of s electrons .(c) Total No. of p- electrons . (d) No. of protons in the nucleus and (e) Valency of the element.

Solution:

The electronic configuration of the element with 2K , 8L and 5M electronic will be

1s^2 2s^2 2p_x^2 2p_y^2 2p_z^2 3s^2 3p_x^1 3p_y^1 3p_z^1

(a) Total no. of electrons = 2+8+5 = 15

therefore  Atomic No. of the element = 15

(b) Total no. of s- electrons = 2+2+2= 6

(c) Total no. of p - electrons = 6+3 = 9

(d) Since the atom is neutral

therefore  No. of protons = No. of electrons = Atomic No. = 15

(e) Since the electron has only three half- filled atomic orbitals , therefore valency of the element = 3