# 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 = ….)

mv

^{2}= ev =

= =

__: Calculate wavelength of spectral line in spwectra of Li__

**Illustration**^{2+}ion when transition takes place b/w two levels whose sum is 6 & difference is 2.

Ans: Let transition takes place b/w levels n

_{1}& n

_{2}

n

_{1}+ n

_{2}= 6 & n

_{2}- n

_{1}= 2

On solving n

_{2}= 4 n

_{2}= 2

= = Rz

^{2}

Here z = 3 (Li

^{2+}ion)

= 109,677 cm

^{-1}x 3

^{2}

= 109,677 x 9

= 109,677 x

=

= 5.4 x 10

^{-6}

__: Calculate (i) I__

**Illustration**^{st}excitation energy yo electron of He

^{+}atm.

(ii) Ionization energy of He

^{+}atom.

atom,

E

_{n}= -

__: What is I__

**Dumb Question**^{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 = E

_{2}- E

_{1}

z = 2

E

_{1}= & E

_{2}=

= - 1.312 x 10

^{6}+ 4 x (1.312 x

^{6}) = 3 x 1.312 x 10

^{2}J/mol

= 3.936 x 10

^{6}J/mol

(ii)

__: What is Ionization energy ?__

**Dumb Question**Ans: Energy required to remove electron from n = 1 to n = i.e.

I.E. = E = - E

_{1}= 0 - (- 1.312 x 10

^{6}x 4)

= 5.248 x 10

^{6}J/mol

__: Why = 0 ?__

**Dumb Question**Ans: E

_{n}& when n

then 0

So, = 0

__….. with electron (m = 9.1 x 10__

**Question**^{-31}kg) moving with velocity 10

^{3}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.

mv

^{2}= ev

x 9.1 x 10

^{-31}x 10

^{6}= 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.

__: Why electron can not exist in nucleus ?__

**Dumb Question**Ans: Diameter of atomic nucleus is order of 10

^{-15}m.

So, max. uncertainity in its position would be 10

^{-15}m (i.e. x = 10

^{-15}m)

Mass of electron = 9.1 x 10

^{-31}

By uncertainity Principle

x.(mv) = v =

v =

= 5.77 x 10

^{10}m/s.

This value is much higher than velocity of light (3 x 10

^{8}m/s). So, it is not possible.

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

**Question**Ans: v = x 500 = 0.1 m/s

x.v =

x = = 5.77 x 10

^{-4}m