Atomic Structure - 6

Rule for finding Group No.:

(i) If last shell contains 1 or 2 electrons, then group no. is 1 & 2 respectively.

(ii) If last shell contains more than 2 elctrons in last shell plus.

(iii) If electrons are present in (n n - 1)d orbital in addition to ns orbital, then gp. No. is equal to total no. of electrons present in (n - 1)d orbital & ns orbital.


Q.1. Two particles A & B are in motion. If wavelength associate with particle A is 5 x 10^-8 m. Caclulate wavelength of B if momentum is half of A.

Ans:  
           But   P_B = P_{A
            B} = 2 <sub>A

                 B</sub> = 10^-7 m


Q.2. Show that wavelength of a moving particle is related to its K.E. (E) as =

Ans: Acording de Broglie   =
But mv² = E   v =



Q.3. A 25 watt bulb emits mouochromatic yellow light of wavelength of 0.57 µm. Calculate rate of emission of quanta/second.

Ans: Energy emitted/sec = 25 J/s
Energy of one photon (E) = h = h
                                   = = 3.48 x 10^-19 J
No. of photons enitted per sec =
                                              = 7.18 x 10¹⁹


Question: Calculate energy required to …….. from n = 2 orbit. What is longest wavelength of light in cm. can used ?

Ans:   E = - E₂ = 0 - = 5.45 x 10^-19 J/atom.
         E = h =
              = 3.647 x 10^-7


Q.5. An ion with mass no. 56 contains 30 mts +ve charge & 30.4% more neutrons than electrons. What is symbol of ion ?

Ans: Suppose no. of electrons in ion M³⁺ = x
No. of neutrons = x + x = 1.304 x
No. of electrons in neutral atom = x + 3
No. of protons = x + 3
Mass no. = No. of photons + No. of neutrons
        56 = x + 3 + 1.304 x
     x = 23
No. of protons = 23 + 3 = 26
Symbol of will be


Q.6. In photon of wavelength 150p, strikes an atom & an bound electron ejected out with velocity 1.5 x 10⁷ m/s. Calculate energy with which it bound to nucleus.

Ans:   Energy of incident photon =
                                              = 13.25 x 10^-16 J
                                             = (9.11 x 10^-31) x 1.5 x 10⁷
                                             = 1.025 x 10^-16 J
Energy with which electron was bound
            = 13.25 x 10¹⁶ J - 1.025 x 10^-16 J = 12.225 x 10^-16 J


Q.7. If uncertainities in measurement of position and momentum of electron are equal in magnitude, what is uncertainity in measurement of velocity.

Ans:   x x p =
As     x = p   (p)² =
        p =
        
           = 7.98 x 10¹² m/s   which is absurd.


Q.8. A bulb emits light of wavelength 4500 A⁰. Bulb is rated as 150 watt & 8% of energy is emitted as light. How many photons are emitted by bulb per second.

Ans: Energy emitted/sec = 150 J/s
Energy emitted as light = x 150 = 12 J = E
   E = nh = nh
   n =
      = 2.717 x 10¹⁹


Question: If H₂ is exposed to radiation of wavelength 253 …, waht % of radiant energy will be converted into K.E. ?

Ans: Energy required to break H-H bond =
                                                        = 7.83 x 10^-19 J
Energy of photon =
                        = 7.83 x 10^-19 J
Energy left after dissociation = K.E. = (7.83 - 7.15) x 10^-19
% of energy used in K.E. = x 100 = 8.68 %


Q.10. Find velocity (m/s) of electron in I^st Bohr orbit of radius a₀. Find be Broglie wavelength (in m). Find angular momentum of 2p orbital of H-atom.

Ans: For H & H-like atoms,
           v_n = 2.188 x 10⁶ x m/s
       For H-atom, z =1   & for I^st orbit   n = 1
       v = 2.188 x 10⁶
       Debroglie wavelength, =
                                          = 3.33 x 10^-10 m
       Orbital angular momentum =
       For   2p, = 1,
       Orbital angular momentum =
                                             =


Q.1. A photon of = 4 x m strikes on metal surface, waork function of metal being 2.13 ev. Calculate (i) energy of photon (ev) (ii) K.E. of electron (iii) velocity of photoelectron.

Ans: (i) Energy of photon = h
                                    = 4.97 x 10^-19 J
       1ev = 1.6 x 10^-19 J
             = 3.1 ev

(ii) K.E. of electrons = h - h₀ = 3.1 - 2.13 = 0.97 ev

(iii) mv² = 0.97 ev = 0.97 x 1.6 x 10^-19
      v² = 0.341 x 10¹²   v = 5.84 x 10⁵ m/s


Q.2. In H-atom, energy of electron is n^th orbit is given   E_n = - ev. Show that
E_{n + 1} - E_n = ev for large values of n.

Ans:   E_{(n + 1)} - E_n =
                         
     E_{n + 1} - E_n
As  n , 0
     E_{n + 1} - E_n =