Atomic Structure - 3

Calculation of de broglie wavelength of electron from potential applied to accelerate it:

If accelerating potential v is applied,
Energy by electron = ev (charge x potential = ….)
    mv² = ev   =
= =


Illustration: Calculate wavelength of spectral line in spwectra of Li²⁺ ion when transition takes place b/w two levels whose sum is 6 & difference is 2.

Ans: Let transition takes place b/w levels n₁ & n₂
      n₁ + n₂ = 6   &   n₂ - n₁ = 2
On solving   n₂ = 4   n₂ = 2
       = = Rz²
Here   z = 3 (Li²⁺ ion)
       = 109,677 cm^-1 x 3²
          = 109,677 x 9
          = 109,677 x
       =
          = 5.4 x 10^-6


Illustration: Calculate (i) I^st excitation energy yo electron of He⁺ atm.
(ii) Ionization energy of He⁺ atom.

atom,
         E_n = -


Dumb Question: What is I^st excitation energy ?

Ans: It is amount of energy required to excite electron from h = 1 (ground state) to     n = 2 (I^st excite state)
       E = E₂ - E₁
       z = 2
       E₁ =   &   E₂ =
           = - 1.312 x 10⁶ + 4 x (1.312 x ⁶) = 3 x 1.312 x 10² J/mol

           = 3.936 x 10⁶ J/mol

(ii) Dumb Question: What is Ionization energy ?
     Ans: Energy required to remove electron from n = 1   to   n =   i.e.
      I.E. = E = - E₁ = 0 - (- 1.312 x 10⁶ x 4)
                                 = 5.248 x 10⁶ J/mol


Dumb Question: Why = 0 ?

Ans: E_n   &   when n
then   0
So, = 0


Question ….. with electron (m = 9.1 x 10^-31 kg) moving with velocity 10³ m/s & find refarding potential required to stop electron ?

Ans:   = = = 7.25 x 10^-7 m
Let v be retarding potential in volt.
To stop electron, K.E> of electron = opposing energy of potential.
       mv² = ev
       x 9.1 x 10^-31 x 10⁶ = 1.6 x 10^-19 x v
       v = volt      v = 2.844 x 10^-6 volt


Derivation of Bohr’s Postulate of angular momentum from de boglie wavelength:

According to De Broglie electron is not only particle but has wave character.
So, circumference of orbit must be equal to integral nultiple of wavelength () for com[letely in phase.
i.e.   2r = n ……………………………… (i)
       r radius of orbit.
But   =
       2r =
    mvr =
It is impossible to measure simultaneously position and momentum of a small particle with absolute accuracy.
Product of uncertainity in momentum (P = mv) & position is constant & is equal to or greater than , where h Planck’s constant.
       x.P   x.(mv)
       x.v
It has significance only for microscopic particles.


Dumb Question: Why electron can not exist in nucleus ?

Ans: Diameter of atomic nucleus is order of 10^-15m.
So, max. uncertainity in its position would be 10^-15m (i.e. x = 10^-15m)
Mass of electron = 9.1 x 10^-31
By uncertainity Principle
       x.(mv) =   v =
       v =
           = 5.77 x 10¹⁰ m/s.
This value is much higher than velocity of light (3 x 10⁸ m/s). So, it is not possible.


Question: ……… with uncertainity of 0.02%. What is uncertainity in position.

Ans:   v = x 500 = 0.1 m/s
         x.v =
      x = = 5.77 x 10^-4 m