Structure Of Atom Important Questions

(i) $1s^2$
(ii)$1s^2$, $2s^2$,$2p^6$
(iii)$1s^2$, $2s^2$,$2p^6$
(iv) $1s^2$
(v) $1s^2$, $2s^2$,$2p^6$
(vi) $1s^2$, $2s^2$,$2p^6$

(i) 2p
(ii) 4p
(iii) 4d
(iv) 3p
(v) 2s
(vi) 4f

(i) represent 4s
(ii) represent 3d
(iii) represent 3p
(iv) represent 3s
The decreasing order of energy of these orbitals is
3d > 4s > 3p > 3s
(ii) > (i) > (iii) > (iv)

n=3 and l=2 represent 3d.
Now possible orbitals are
m_l=-2,-1,0, +1, +2

(i) Number of electrons=6
Number Of protons=6
Mass number =12
Number of Neutrons= 12 -6 =6
(ii) Number of electrons=6
Number Of protons=6
Mass number =14
Number of Neutrons= 14 -6 =8

Using
$\lambda = \frac {h}{mV}$
or
$V= \frac {h}{m \lambda}$
Subsituting the values
V=.1516 × 10⁹ m/s
Kinetic energy = $\frac {1}{2}mV^2 $
Subtituting the value of V and m
K.E=1.047 × 10^-14 J

Using
$\frac {1}{\lambda} = R_H Z^2 (\fracc{1}{n_1^2} - \fracc{1}{n_2^2})$
Substituting the values $n_1 =2$ and $n_2=4$, Z=1
$\lambda =.2667 \times 10^{-7}$ m
Now energy is given by
$E= \frac {hc}{\lambda} = 74.47 \times 10^{-20} $ J/atom

Using de-broglie relation
$\lambda = \frac {h}{p} = \frac {h}{mV}$
Substituting the values
$\lambda = 6.626 \times 10^{-34} $ m

Now
$K= \frac {1}{2}mV^2$
or
V = \sqrt {\frac {2K}{m}}$
Subsituting the values
V=.7 × 10³ m/s
Now De-Broglie wavelength is given by
$\lambda = \frac {h}{p} = \frac {h}{mV}$
Substituting the values
$\lambda = 1.036 \times 10^{-6} $ m

(i) $E= h \nu= \frac {hc}{\lambda}$
Substituting the values
$E=3.1 eV$
(ii) From Photo-electric effect equation
K= E - W= 3.1 - 2.13 =.97 eV
(iii) Now
$\frac {1}{2}mV^2 =.97 \times 1.6020 \times 10^{-19}$
Substituting the values
v=5.84 × 10⁵ m/s

According to Heisenberg's uncertainty principle,
$\Delta x \Delta p \ge \frac {h}{2 \pi}$
Here $\Delta x =1 \times 10^{-10}$ m
Plank's Constant (h)=6.63 × 10^-34 J-s
Therefore uncertainty in the momentum of an electron will be
$\Delta p \ge \frac {h}{2 \pi \Delta x}$
or
$\Delta p \ge 5.27 \times 10^{-25} \ kgms^-1$
Therefore uncertainty in the momentum of an electron =$5.27 \times 10^{-25} \ kgms^-1$

The energy of the electron in the nth shell of hydrogen atom is given by
$E_n= - \frac {2 \pi ^2 me^4}{n^2 h^2}= - \frac {1.312 \times 10^6}{n^2} J/mol$
(i) First excitation energy is the amount of energy required to excite the electron from n = 1 (ground state) to n = 2 (first excited state)
$\Delta E = E_2 - E_1 $
$E_2= - \frac {1.312 \times 10^6}{2^2}$
$E_1= - \frac {1.312 \times 10^6}{1^2}$
Therefore
$\Delta E = 9.84 \times 10^5 $ J/mol
(ii) Ionization energy is the amount of energy required to remove the electron from n=1 to $n=\infty$
$\Delta E = E_{\infty} - E_1 $
$E_{\infty}= - \frac {1.312 \times 10^6}{(\infty)^2}=0$
$E_1= - \frac {1.312 \times 10^6}{1^2}$
Therefore
$\Delta E = 1.312 \times 10^6 $ J/mol

Radius is given by
$r_n= n^2 \times .529 \times 10^{-10} $ m
$r_3= 3^2 \times .529 \times 10^{-10}=4.761 \times 10^{-10} \ m$
Now
$mvr= \frac {nh}{2 \pi}$
or
$v= \frac {nh}{2 \pi m r}$
Substituting the values
$v=7.29 \times 10^5$ m/s
Number of Revolutions per sec
$= \frac {v}{2 \pi r}$
Substituting the values
$=2.4 \times 10^{14} $ rev/sec

Ionization energy will be equal to energy possessed by the Electromagnetic radiation
$E= \frac {hc}{\lambda}$
Substituting the values
$E=8.21 \times 10^{-19}$ J
Now This is the energy per atom
Energy per Mole
$E=8.21 \times 10^{-19} \times 6.023 \times 10^{23} = 494000$ J = 494 KJ

(i)1
(ii) 0
(iii) 2
(iv) 0

I.
(a) 1s < 2s < 2p < 3s
(b) 3s < 3p < 4s < 4d
(c) 4d < 5p < 6s < 4f < 5d
(d) 7s < 5f < 6d < 7p

II. (a) 5s (b) 5f