atomic

1. The typical wavelengths emitted by diatomic molecules in purely vibrational and purely rotational transitions are respectively in the region of
1. infrared and visible
2. visible and infrared
3. infrared and microwaves
4. microwaaves and infrared
(C) infrared and microwaves
2. In $^3S$ state of the Helium atom, the possible values of the total electronic angular momentum quantum numbers are:
   1. 0 (zero) only
   2. 1 only
   3. 0,1 and 2
   4. 0 and 1 only
   The term symbol is given by $^{(2s+1)}L_J$. For $^3S$ state $(2s+1)=3$, hence $s=1$. For $S$ state $L=0$. Hence $J=L+S=1$. Hence answer is (B)1 only
3. The outer electron configuration of divalent Manganese ion is $3d^54s^0$. The ground state of this ion is characterized by the spectroscopic term:
1. $^6S_{5/2}$
2. $^2D_{5/2}$
3. $^2F_{5/2}$
4. $^6H_{5/2}$
The rules governing the term symbol for the ground state according to L-S coupling scheme are given below:
1. The spin multiplicity is maximized i.e., the electrons occupy degenerate orbitals so as to retain parallel spins as long as possible (Hund’s rule).
2. The orbital angular momentum is also maximized i.e., the orbitals are filled with highest positive m values first.
3. If the sub-shell is less than half-filled, $J = L– S$ and if the sub-shell is more than half – filled, $J = L +S$.
The term symbol is given by $^{2S+1}L_J$. The left-hand superscript of the term is the spin multiplicity, given by $2S+1$ and the right- hand subscript is given by $J$.
For $d^5$ configuration we have
	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$
$m_l=$	+2	+1	0	-1	-2
Hence, \begin{align*} L&={\scriptstyle(+2)+(+1)+(0)+(-1)+(-2)}\\ &=0 \end{align*} i.e. $S$ state \begin{align*} S&={\scriptstyle(+1/2)+(+1/2)+(+1/2)+(+1/2)+(+1/2)}\\ &=5/2 \end{align*} $$2S+1=6$$ $$J=L+S=0+5/2=5/2$$ Term symbol is $^6S_{5/2}$
4. Consider a nuclear $F^{19}$. When it is placed in a magnetic field of 1.0 tesla, the resonance frequency (in units of MHz) of the signal observed for this nucleus in the NMR spectrometer is : (Given: $g_N=5.256$, $\mu_N=5.0504\times10^{-27} J/T$; the subscript $N$ refers to the nuclear factors)
1. 30 MHz
2. 90 MHz
3. 40 MHz
4. 5.0 MHz
The resonance condition for NMR is $$h\nu=g_N\mu_NB_0$$ \begin{align*} \nu&=\frac{g_N\mu_NB_0}{h}\\ &=\frac{5.256\times5.0504\times10^{-27}\times1}{6.626\times10^{-34}}\\ &=40~MHz \end{align*}
5. The 623.8 nm radiation emitted by a He-Ne laser due to the transition between :
1. $3s$ and $2p$ levels of Ne
2. $3s$ and $3p$ levels of Ne
3. $2p$ and $2s$ levels of Ne
4. $2p$ and $1s$ levels of Ne
(A)$3s$ and $2p$ levels of Ne
6. In the Stern-Gerlach experiment, the number of components in which the atomic beam splits depends upon the value of:
1. $l$
2. $s$
3. $j$
4. $m_j$
(C)$j$
7. The Lande's splitting factor for the atomic state $^2P_{3/2}$ is :
1. $1/3$
2. $2/3$
3. $1$
4. $4/3$
For $^2P_{3/2}$ state $(2s+1)=2$, hence $s=1/2$, $j=3/2$, for $P$ state $l=1$ \begin{align*} g&={\scriptstyle1+\frac{j(j+1)+s(s+1)-l(l+1)}{2j(j+1)}}\\ &={\scriptstyle1+\frac{\frac{3}{2}(\frac{3}{2}+1)+\frac{1}{2}(\frac{1}{2}+1)-1(1+1)}{2\frac{3}{2}(\frac{3}{2}+1)}}\\ &=\frac{4}{3} \end{align*}
8. The "Normal" and "Anomalous" Zeeman effects are observed when (here S' is the total spin angular momentum due to the coupling of individual spin angular momenta)
1. $S'=0$ and $S'\ne0$, respectively
2. $S'=0$ and $S'=0$, respectively
3. $S'\ne0$ and $S'=0$, respectively
4. $S'\ne0$ and $S'\ne0$, respectively
(A)$S'=0$ and $S'\ne0$, respectively
9. The number of photons emitted per second from 1 watt Ar-ion laser operating at 488 nm is approximately
1. $10.23\times10^{19}$
2. $2.46\times10^{18}$
3. $10.23\times10^{17}$
4. $2.46\times10^{15}$
$Power=P=1 watt$, $\lambda=488 ~nm=488\times10^{-9}m$ $$P=NE=Nh\nu=N\frac{hc}{\lambda}$$ \begin{align*} N&=P\frac{\lambda}{hc}\\ &=1\times\frac{488\times10^{-9}}{6.63\times10^{-34}\times3\times10^8}\\ &=2.46\times10^{18} \end{align*}
10. For a diatomic molecule with the vibrational quantum number $n$ and rotational quantum number $J$, the vibrational level spacing $\Delta E_n=E_n-E_{n-1}$ and the rotational level spacing $\Delta E_J=E_J-E_{J-1}$ are approximately
1. $\Delta E_n=$ constant, $\Delta E_J=$ constant
2. $\Delta E_n=$ constant, $\Delta E_J\propto J$
3. $\Delta E_n\propto n$, $\Delta E_J\propto J$
4. $\Delta E_n\propto n$, $\Delta E_J\propto J^2$
(B)$\Delta E_n=$ constant, $\Delta E_J\propto J$
11. Consider a hydrogen atom undergoing a $2P\rightarrow 1S$ transition. The lifetime $t_{sp}$ of the $2P$ state for spontaneous emission is $1.6 ns$ and the energy difference between the levels is $10.2eV$. Assuming that the refractive index of the medium $n_0 = 1$, the ratio of Einstein coefficients for stimulated and spontaneous emission $B_{21}(\omega)/A_{21}(\omega)$ is given by
1. $0.683\times10^{12} m^3J^{-1}s^{-1}$
2. $0.146\times10^{-12} Jsm^{-3}$
3. $6.83\times10^{12} m^3J^{-1}s^{-1}$
4. $1.46\times10^{-12} Jsm^{-3}$
\begin{align*} \frac{B_{21}(\omega)}{A_{21}(\omega)}&=\frac{\pi^2c^3}{\hbar\omega^3n_0^3}\\ &=\frac{\pi^2c^3\hbar^2}{(\Delta E)^3n_0^3}\\ &=\frac{c^3h^2}{4(\Delta E)^3n_0^3} \end{align*} $\Delta E=10.2 eV$, $n_0=1$ \begin{eqnarray} \begin{array}{c}\frac{B_{21}(\omega)}{A_{21}(\omega)}=\frac{\left(3\times10^8\right)^3\times\left(6.63\times10^{-34}\right)^2}{\left(10.2\times1.6\times10^{-19}\right)^3\times1}\\ =0.683\times10^{12}m^3J^{-1}s^{-1}\end{array} \end{eqnarray} Hence, answer is (A)
12. The LS configurations of the ground state of $^{12}Mg$, $^{13}Al$, $^{17}Cl$ and $^{18}Ar$ are, respectively,
1. $^3S_1$, $^2P_{1/2}$, $^2P_{1/2}$ and $^1S_{0}$
2. $^3S_1$, $^2P_{3/2}$, $^2P_{3/2}$ and $^3S_{1}$
3. $^1S_0$, $^2P_{1/2}$, $^2P_{3/2}$ and $^1S_{0}$
4. $^1S_0$, $^2P_{3/2}$, $^2P_{1/2}$ and $^3S_{1}$

Electronic configuration of $^{12}Mg$ is $1s^22s^22p^63s^2$

There are two 3s electrons with $s_1=\frac{1}{2}$ and $s_2=\frac{1}{2}$. Hence, $S=1$ and $2S+1=3$

For two 3s electrons each with $l=0$, $m_l=0$. Hence, $L=0$

$J=L+S=0$. Hence, term symbol for $^{12}Mg$ is $^1S_0$

For, $^{13}Al$, electronic configuration is $1s^22s^22p^63s^23p^1$

$S=\frac{1}{2}$, $2S+1=2$, $L=1$, $J=L-S=1-\frac{1}{2}=\frac{1}{2}$ (less than half-filled)

Hence, term symbol for $^{13}Al$ is $^2P_{1/2}$

Hence, answer is (C).

13. There is no infrared absorption for nitrogen molecule because:
1. its polarizability is zero
2. it has no vibrational levels
3. it has no rotational levels
4. its dipole moment is zero
According to selection rule of IR absorption there should be change in dipole moment of the molecule. Nitrogen molecule is homonuclear diatomic molecule. It has no dipole moment and also there is no change in its dipole moment during vibration.

Hence answer is (D)

14. In the first order Stark effect in hydrogen atom the ground state split into :
1. 2 levels
2. 3 levels
3. 4 levels
4. does not split
Splitting of spectral lines in the presence of external electric field is called a Stark effect. The first order correction to the ground state energy of hydrogen atom in the presence of electric $\bar\epsilon$ field applied in z-direction is given by \begin{align*} \Delta E_0&=\int\psi^*_{100}H'\psi_{100}d\tau\\ &=\int\psi^*_{100}(e\epsilon z)\psi_{100}d\tau\\ &=e\epsilon\int z\left|\psi_{100}\right|^2d\tau\\ &=0 \end{align*} Since, integrand is odd function. Hence, there is no splitting in the ground state energy.

Hence, answer is (D)

15. According to Hund's rule, the ground state of Si (atomic number 14) atom is
1. $^1P_1$
2. $^3P_0$
3. $^3P_3$
4. $^3D_3$
The electron configuration of Si is $1s^22s^22p^63s^23p^2$. Hence, there are 2 unpaired electrons in $p$ state. Hence, total spin is $S=\frac{1}{2}+\frac{1}{2}=1$. Multiplicity is $(2S+1)=3$.

For $3p^2$ electrons $m_{l_1}=+2$, $m_{l_2}=+1$, hence $L=2+1=3$

Hence, $J=L-S=0$

term symbol is $^3P_0$

Hence, answer is (B)

16. A source with bandwidth of $10^{-3}\:nm$ centered about $\lambda=500\:nm$ has a coherence length
1. 0.25 m
2. 2.5 $\mu$m
3. 25 cm
4. 2.5 m
Coherence length \begin{align*} l_c&=\frac{\lambda^2}{\Delta\lambda}\\ &=\frac{25\times10^{4}}{10^{-3}}\\ &=25\times10^{7} nm=0.25 m \end{align*}
17. For a two level system, the population of atoms in the upper and lower levels are $3\times10^{18}$ and $0.7\times10^{18}$, respectively. If the coefficient of stimulated emission is $3\times10^{5}\:m^3/W-s^3$ and the energy density is $9.0 J/m^3Hz$, the rate of stimulated emission will be
1. $6.3\times10^{16}\:s^{-1}$
2. $4.1\times10^{16}\:s^{-1}$
3. $2.7\times10^{16}\:s^{-1}$
4. $1.8\times10^{16}\:s^{-1}$

Rate of stimulated emission is $$R_{stimu}=\rho B$$ $$R_{stimu}= 9.0 \times 3\times10^{5}$$ $$R_{stimu}=2.7\times10^{16}\:s^{-1}$$

Hence, answer is (C)

18. The output of a laser has a pulse width of 30 ms and average output power of 0.6 watt per pulse. If the wavelength of the laser light is 640 nm. How many photon does each pulse contain?
1. $2.9\times10^{18}$
2. $3.5\times10^{18}$
3. $5.8\times10^{15}$
4. $6.5\times10^{16}$

$\Delta t=30 ms=30\times10^{-3}s$, $P_{avg}=0.6 \text{watt per pulse}$, $\lambda=640 nm=640\times10^{-9}m$ Energy per pulse is power times pulse duration. \begin{align*} E_{pulse}&=P_{avg}\Delta t\\ &=0.6\times30\times10^{-3}\\ &=0.018 joule \end{align*} Energy per photon is given by Planck formula. \begin{align*} E_{photon}&=h\nu=\frac{hc}{\lambda}\\ &=\frac{6.63\times10^{-34}\times3\times10^8}{640\times10^{-9}}\\ &=0.031078125\times10^{-17} joule \end{align*} Number of photons per pulse is just energy per pulse divided by energy per photon \begin{align*} N&=\frac{E_{pulse}}{E_{photon}}\\ &=\frac{0.018}{0.031078125\times10^{-17}}=5.8\times10^{16} \end{align*}

Hence, answer is (C)

19. The number of fundamental vibrational modes of $CO_2$ molecule is :
1. 4 : 2 Raman active and 2 IR active
2. 4 : 1 Raman active and 3 IR active
3. 3 : 1 Raman active and 2 IR active
4. 3 : 2 Raman active and 1 IR active

$CO_2$ molecule is a linear molecule. Hence, number of vibrational modes are given by $3N-5$. $CO_2$ is triatomic molecule, hence $N=3$. Hence, number of modes $=3\times 3-5=4$. Out of which 1 is Raman active and 3 are IR active.

Hence, answer is (B)

20. The $L$, $S$ and $J$ quantum numbers corresponding to the ground state electronic configuration of Boron $(z = 5)$ are :
1. $L = 1$, $S = 1/2$, $J = 3/2$
2. $L = 1$, $S = 1/2$, $J = 1/2$
3. $L = 1$, $S = 3/2$, $J = 1/2$
4. $L = 0$, $S = 3/2$, $J = 3/2$

To determine ground state, we will use following Hund's rules

1. Given more than one allowed value for $S$, choose largest possible value.
2. Given more than one allowed value for $L$, choose largest possible value.
3. Given more than one allowed value for $J$, choose largest possible value of $J$ if subshell is more than half-filled and choose smallest possible value of $J$ if subshell is less than half-filled.

The ground state of Boron has a $1s^22s^22p^1$ configuration. Hence, there is single electron in $2p$ state, for which, $S=1/2$, $L=1$. Hence, $J=L-S=1/2$.

Hence, answer is (B)

21. Which of the following molecules will not be sensitive to microwave spectroscopy?
1. $LiH$
2. $CO$
3. $CH_4$
4. $CCl_3$

Microwave spectrum can not be observed for centrosymmetric linear molecules such as $N_2$ (dinitrogen) or HCCH (ethyne), which are non-polar. Tetrahedral molecules such as $CH_4$ (methane), which have both a zero dipole moment and isotropic polarizability, would not have a pure rotation spectrum.

Hence, answer is (C)

22. Light of wavelength 1.5 $\mu m$ incident on a material with a characteristic Raman frequency of $20\times 10^{12}$ Hz results in a Stokes shifted line of wavelength:
1. 1.47 $\mu m$
2. 1.57 $\mu m$
3. 1.67 $\mu m$
4. 1.77 $\mu m$

Raman shift in terms of wavelength is given by $$\frac{1}{\Delta\lambda}=\frac{1}{\lambda_0}-\frac{1}{\lambda_1}$$ $$\Delta\lambda=\frac{\lambda_0\lambda_1}{\lambda_1-\lambda_0}$$ where, $\lambda_0$ is wavelength of incident light, $\lambda_1$ is the Raman spectrum wavelength. Here, $\lambda_0=1.5\mu m$ $\lambda_1=\frac{c}{\nu}=\frac{3\times10^8}{20\times 10^{12}}=15\mu m$ $$\Delta\lambda=\frac{1.5\times15}{15-1.5}=1.67\mu m$$

Hence, answer is (C)

23. The far infrared rotational absorption spectrum of a diatomic molecule shows equidistant lines with a spacing 20 cm$^{-1}$. The position of the first Stokes line in the rotational Raman spectrum of this molecule is:
1. 20 cm$^{-1}$
2. 40 cm$^{-1}$
3. 60 cm$^{-1}$
4. 120 cm$^{-1}$

The energies of the rotational levels are given by $$E_J=J(J+1)\frac{\hbar^2}{2I}$$ The transition energies for absorption of radiation are given by \begin{align*} h\nu&=E_f-E_i\\ &=J_f(J_f+1)\frac{\hbar^2}{2I}-J_i(J_i+1)\frac{\hbar^2}{2I} \end{align*} But selection rule for transition is $\Delta J=\pm1$. Hence, $J_f=J_i+1$. $$\nu=2(J_i+1)\frac{\hbar^2}{2Ih}$$ In terms of wavenumber $$\bar\nu=2(J_i+1)\frac{\hbar^2}{2Ihc}$$ $$\bar\nu=2B(J_i+1)$$ where, $B=\frac{\hbar^2}{2Ihc}$

The lowest energy transition is between $J_i = 0$ and $J_f = 1$ so the first line in the spectrum appears at a frequency of $2B$. The next transition is from $J_i = 1$ to $J_f = 2$ so the second line appears at $4B$. The spacing of these two lines is $2B$. In fact the spacing of all the lines is $2B$ according to this equation. $$2B=20\: cm^{-1}\Rightarrow B=10$$ Stokes lines are observed at $$\bar\nu=\bar\nu_0-B(4J+6)\:cm^{-1}$$ Hence, there is a gap of $6B$ between $\bar\nu_0$ and $1^{st}$ Stokes line. $$\bar\nu_0-\bar\nu=6B=60\:cm^{-1}$$

Hence, answer is (C)

24. The trivalent gadolinimum ion has seven electrons in its outer orbital. The Lande $g$ factor for this ion is:
1. 1
2. $\frac{3}{2}$
3. 2
4. $\frac{5}{2}$

The trivalent gadolinimum has electron configuration $[Xe]4f^7$. Hence, there are 7 electrons in $f$ orbital. For $f^7$ configuration we have

	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$	$\uparrow$
$m_l=$	+3	+2	+1	0	-1	-2	-3

Hence, \begin{align*} L&={\scriptstyle(+3)+(+2)+(+1)+(0)+(-1)+(-2)+(-3)}\\ &=0 \end{align*} i.e. $S$ state \begin{align*} S&={\scriptstyle\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}}\\ &=7/2 \end{align*} $$2S+1=8$$ $$J=L+S=0+7/2=7/2$$ Term symbol is $^8S_{7/2}$ \begin{align*} g_J&=1+\frac{J(J+1)+S(S+1)-L(L+1)}{2J(J+1)}\\ &=1+\frac{\frac{7}{2}(\frac{7}{2}+1)+\frac{7}{2}(\frac{7}{2}+1)-0}{2\frac{7}{2}(\frac{7}{2}+1)}\\ &=2 \end{align*}

Hence, answer is (C)

25. Of the following term symbols of the $np^2$ atomic configurations, $^1S_0$, $^3P_0$, $^3P_1$, $^3P_2$ and $^1D_2$, which is the ground state?
1. $^3P_0$
2. $^1S_0$
3. $^3P_2$
4. $^3P_1$

According to Hund’s rules

1. State with highest multiplicity has lowest energy
2. State with same multiplicity, the state with highest $L$ will have lowest energy
3. State with same multiplicity and $L$ value. The state with lowest $J$ has lowest energy (only if subshell is less than half filled)

From the given states $^1S_0$, $^3P_0$, $^3P_1$, $^3P_2$ and $^1D_2$, $^3P_0$ will have the lowest energy

Hence, answer is (A)

26. A diatomic molecule has vibrational states with energies $E_\nu = \hbar\omega \left(\nu+\frac{1}{2}\right)$ and rotational states with energies $E_j = Bj(j + 1)$, where $\nu$ and $j$ are non-negative integers. Consider the transitions in which both the initial and final states are restricted to $\nu\leq 1$ and $j\leq 2$ and subject to the selection rules $\Delta\nu = \pm 1$ and $\Delta j = \pm 1$. Then the largest allowed energy of transition is
1. $\hbar\omega-3B$
2. $\hbar\omega-B$
3. $\hbar\omega+4B$
4. $2\hbar\omega+B$

$$E=\hbar\omega \left(\nu+\frac{1}{2}\right)+Bj(j + 1)$$ where $j$ is lower quantum number For $\Delta\nu = \pm 1$ and $\Delta j = \pm 1$ $$\Delta E=E_i-E_f=\hbar\omega+2B(j + 1)$$ For the largest allowed energy of transition $j=1$ $$\Delta E=E_i-E_f=\hbar\omega+4B$$

Hence, answer is (C)

27. A He-Ne laser operates by using two energy levels of Ne separated by 2.26 eV. Under steady state conditions of optical pumping, the equivalent temperatures of the system at which the ratio of the number of atoms in the upper state to that in the lower state will be 1/20, is approximately (the Boltzman constant $k_B=8.6\times10^{-5}\:eV/K$
1. $10^{10}\:K$
2. $10^{8}\:K$
3. $10^{6}\:K$
4. $10^{4}\:K$

$$\frac{N_2}{N_1}=e^{-E/k_BT}$$ \begin{align*} T&=-\frac{E}{k_B}\frac{1}{\ln{\frac{N_2}{N_1}}}\\ &=-\frac{2.26}{8.6\times10^{-5}}\frac{1}{\ln{\frac{1}{20}}}\\ &\approx 10^{4} \end{align*}

Hence, answer is (D)

28. In the microwave spectrum of identical rigid diatomic molecules, the separation between the spectral lines is recorded to be $0. 7143\: cm^{-1}$. The moment of inertia of the molecule, in $kg\: m^2$, is
1. $2.3 \times 10^{-36}$
2. $2.3 \times 10^{-40}$
3. $7.8 \times 10^{-42}$
4. $7.8 \times 10^{-46}$

In rotational spectra separation between two spectral line is $2B=0. 7143\: cm^{-1}$ $$B=\frac{h}{8\pi^2cI}$$ When $B$ is in $cm^{-1}$, $c$ should be taken in $cm\:s^{-1}$. \begin{align*} I&=\frac{h}{4\pi^2c2B}\\ &=\frac{6.62607\times10^{-34}}{4\times3.14^2\times3\times10^{10}\times0.7143}\\ &=7.832\times10^{-46} \end{align*}

Hence, answer is (D)

29. Which one of the following electronic transitions in neon is NOT responsible for LASER action in a helium-neon laser?
1. $6s\rightarrow5p$
2. $5s\rightarrow4p$
3. $5s\rightarrow3p$
4. $4s\rightarrow3p$

$6s\rightarrow5p$

Hence, answer is (A)

30. In the linear Stark effect, the application of an electric field
1. completely lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into four levels
2. partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into three levels
3. partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into two levels
4. does not affect the $n= 2$ levels

Partially lifts the degeneracy of $n = 2$ level of hydrogen atom and splits $n = 2$ level into three levels

Hence, answer is (B)

31. In hyperfine interaction, there is coupling between the electron angular momentum $\vec J$ and nuclear angular momentum $\vec I$, forming resultant angular momentum $\vec F$. The selection rules for the corresponding quantum number $F$ in hyperfine transitions are
1. $\Delta F=\pm2$ only
2. $\Delta F=\pm1$ only
3. $\Delta F=0,\pm1$
4. $\Delta F=\pm1,\pm2$

$\Delta F=0,\pm1$

Hence, answer is (C)

32. A vibrational-electronic spectrum of homonuclear binary molecules, involving electronic ground state $\epsilon''$ and excited state $\epsilon'$, exhibits a continuum at $\bar\nu cm^{-1}$. If the total energy of the dissociated atoms in the excited state exceeds the total energy of the dissociated atoms in the ground state by $E_{ex} cm^{-1}$, the dissociation energy of the molecule in the ground state is
1. $(\bar\nu+E_{ex})/2$
2. $(\bar\nu-E_{ex})/2$
3. $(\bar\nu-E_{ex})$
4. $\sqrt{(\bar\nu^2-E_{ex}^2)}$

$D=(\bar\nu-E_{ex})$

Hence, answer is (C)

33. The NMR spectrum of ethanol $(CH_3CH_2OH)$ comprises of three bunches of spectral lines. The number of spectral lines in the bunch corresponding to $CH_2$ group is
1. 1
2. 2
3. 3
4. 4

In ethanol the intensities are in the ratio 3:2:1, because there are three $CH_3$ protons, two $CH_2$ protons and one $OH$ proton in a molecule. So the number of spectral lines in the bunch corresponding to $CH_2$ group is 2

Hence, answer is (B)

34. The ground state electronic configuration of $^{22}Ti$ is $[Ar]3d^54s^2$. Which state, in the standard spectroscopic notations, is not possible in this configuration?
1. $^1F_{3}$
2. $^1S_{0}$
3. $^1D_{2}$
4. $^3P_{0}$

The rules governing the term symbol for the ground state according to L-S coupling scheme are given below:

1. If all shells and subshells are full then the term symbol is $^1S_0$.
2. The spin multiplicity is maximized i.e., the electrons occupy degenerate orbitals so as to retain parallel spins as long as possible (Hund’s rule).
3. The orbital angular momentum is also maximized i.e., the orbitals are filled with highest positive $m_l$ values first.
4. The overall $S$ is calculated by multiplying $\frac{1}{2}$ times the number of unpaired electrons. The overall $L$ is calculated by adding the $m_l$ values for each electron (so if there are two electrons in the same orbital, add twice that orbital's $m_l$).
5. If the sub-shell is less than half-filled, $J = L– S$ and if the sub-shell is more than half – filled, $J = L +S$.

The term symbol is given by $^{2S+1}L_J$. The left-hand superscript of the term is the spin multiplicity, given by $2S+1$ and the right- hand subscript is given by $J$.
The electron configuration of Ti is $1s^22s^22p^63s^23p^63d^24s^2$. For $d^2$ configuration we have

	$\uparrow$	$\uparrow$
$m_l=$	+2	+1	0	-1	-2

Hence, \begin{align*} L&=(+2)+(+1)=3\\ \end{align*} i.e. $F$ state \begin{align*} S&=(+1/2)+(+1/2)=1\\ \end{align*} $$2S+1=3$$ $$J=L-S=3-1=2$$ Term symbol for Ti is $^3F_{2}$

Let us determine which states are possible

If we have

	$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1S_0$ state

If we have

		$\uparrow\downarrow$
$m_l=$	+2	+1	0	-1	-2

We get $^1D_2$ state

If we have

	$\uparrow$			$\uparrow$
$m_l=$	+2	+1	0	-1	-2

We get $^3P_0$ state

For $^1F_3$ state we must have $S=0$, $L=3$ and $J=3$. However, it is not possible to arrange electrons to get $S=0$, $L=3$ and $J=3$.

Hence, answer is (A)

35. The output of a laser has a bandwidth of $1.2\times10^{14}$ Hz. The coherence length $l_c$ of the output radiation is
1. 3.6 mm
2. 50 $\mu$m
3. 2.5 $\mu$m
4. 1.5 cm

Coherence length of laser is given by $$l_c=\frac{c}{\Delta\nu}$$ $$l_c=\frac{3\times10^8}{1.2\times10^{14}}=2.5 \mu m$$

Hence, answer is (C)