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

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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

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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

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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)

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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

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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`

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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)`

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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)`

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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

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