Problem set 73

1. For the logic circuit given below, the decimal count sequence and the modulus of the circuit corresponding to A B C D are (here, J=K=1 locked)
   
1. $8\rightarrow4\rightarrow2\rightarrow1\rightarrow9\rightarrow5$ (mod 6)
2. $8\rightarrow4\rightarrow2\rightarrow9\rightarrow5\rightarrow3$ (mod 6)
3. $2\rightarrow5\rightarrow9\rightarrow1\rightarrow3$ (mod 5)
4. $8\rightarrow5\rightarrow1\rightarrow3\rightarrow7$ (mod 5)

In the T or "toggle" flip-flop both J=1 and k=1 and it changes its output on each clock edge, giving an output which is half the frequency of the signal to the T input.

Clock	A	B	C	D	Decimal
0	1	0	0	0	8
1	0	1	0	0	4
2	0	0	1	0	2
3	1	0	0	1	9
4	0	1	0	1	5
5	0	0	1	1	3
6	1	0	0	0	8

Modulus of a counter, is the number of states that the counter will sequence through before returning back to its original value.

Hence, answer is (B)

2. The low-energy electronic excitations in a two-dimensional sheet of graphene is given by $E(\vec k)=\hbar v k$, where $v$ is the velocity of the excitations. The density of states is proportional to
1. $E$
2. $E^{3/2}$
3. $E^{1/2}$
4. $E^{2}$

In two-dimensions area of k-space is $$\Omega _{2,k}\propto k^2$$ Using $E(\vec k)=\hbar v k$ $$\Omega _{2,k}\propto \frac{E^2}{\hbar^2 v^2}$$ $$D_n(E)=\frac{d\Omega _{n,k}}{dE}$$ $$D_2(E)=\frac{d\Omega _{2,k}}{dE}\propto E$$

Hence, answer is (A)

3. X-ray of wavelength $\lambda=a$ is reflected from $(111)$ plane of a simple cubic lattice. If the lattice constant is $a$, the corresponding Bragg angle (in radian) is
1. $\pi/6$
2. $\pi/4$
3. $\pi/3$
4. $\pi/8$

$$d_{hkl}=\frac{a}{\sqrt{h^2+k^2+l^2}}$$ $$d_{111}=\frac{a}{\sqrt{3}}$$ $$\sin\theta=\frac{\lambda}{2d_{111}}$$ $$\sin\theta=\frac{\sqrt{3}}{2}$$ $$\theta=\pi/3$$

Hence, answer is (C)

4. Let us approximate the nuclear potential in the shell model by a three dimensional isotropic harmonic oscillator. Since the lowest two energy levels have angular momenta $l=0$ and $l=1$ respectively, which of the following two nuclei have magic numbers of protons and neutrons?
1. $_2^4He$ and $_8^{16}O$
2. $_1^2D$ and $_4^{8}Be$
3. $_2^4He$ and $_4^{8}Be$
4. $_2^4He$ and $_6^{12}C$

The energy eigenvalues of the harmonic oscillator are $$E_n=(n+\frac{1}{2})\hbar\omega\quad n=1,2,3,\cdots$$ In case of harmonic oscillator for a given $n$ $$l=n-1,n-2,\cdots0$$ $$m_l=-l,-l+1,\dots l$$ $$m_s=\pm\frac{1}{2}$$ For $l=0$, $n=1$, $m_l=0$ and $m_s=\pm\frac{1}{2}$. Hence there are two states explaining the first magic number, 2. Two nucleons of a given type can occupy the lowest energy level. $_2^4He$ has the lowest energy level for protons completely filled with its two protons, and the lowest level for neutrons completely filled with its two neutrons. That makes $_2^4He$ the first doubly-magic nucleus.

For $l=1$, $n=2$, $m_l=-1, 0,1$ and $m_s=\pm\frac{1}{2}$. Therefore, $l=1$ at $n=2$ corresponds to 6 energy states. Combined with the two $l= 0$ states at energy level $n=1$, that gives a total of 8. The second magic number 8 has been explained! It requires 8 nucleons of a given type to fill the lowest two energy levels. It makes $_8^{16}O$ with 8 protons and 8 neutrons the second doubly-magic nucleus.

Hence, answer is (A)

5. The reaction $^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$ cannot proceed via strong interactions because it violets the conservation of
1. angular momentum
2. electric charge
3. baryon number
4. isospin

A nucleus of deuteron contains one proton and one neutron (each with a baryon number of 1) and has a baryon number of 2 and charge +1. isospin=0, Owing to the generalized Pauli principle, it must have a symmetric spin state, i.e. $S= 1$, and a total angular momentum $J=1$. Helium nucleus is composed of two protons and two neutrons. Hence, Baryon number 4 and charge +2, isospin=0

	$^2_1D+^2_1D\rightarrow ^4_2He+\pi^0$
Q	$1+1\rightarrow 2+0$	Conserved
B	$2+2\rightarrow 4+0$	Conserved
Isospin	$0+0\rightarrow 0+1$	Not Conserved

Hence, answer is (D)