Problem set 3

1. Calculate equivalent temperature of 1eV energy
$E=3/2kT$, $1eV=11604.52500617K$
2. What are unit and dimensions of wavefunction $\psi$
Wavefunction may be in position space or momentum space. For a 1-dimensional position space wavefunction $\psi(x)$ the normalization condition is $\int\psi^*(x)\psi(x)dx=1$, so $\psi^*(x)\psi(x)$ has units of inverse distance and $\psi(x)$ has units of square root of inverse distance, e.g. $m^{−1/2}$. For a 2-dimensional position space wavefunction $\psi(x,y)$ the normalization condition would be $\iint\psi^*(x,y)\psi(x,y)dxdy=1$, so $\psi^*(x,y)\psi(x,y)$ has units of inverse area and $\psi(x,y )$ has units of square root of inverse area, e.g. $m^{−1}$. Similarly, for 3-dimension $m^{-3/2}$.
In momentum space of 1-dimensional case we have $p^{−1/2}$ i.e. $\frac{1}{\sqrt{kg-m}}$
3. 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
4. Solar cell is a type of :
1. Photo-conductive device
2. Photo-emissive device
3. Photo-voltaic device
4. Electromagnetic device
(C) Photo-voltaic device.
5. KCL and KBr are alkali halides, both having the NaCl crystal structure. However, in the X-ray diffraction certain reflections are absent for KCl as compared to KBr, for example (111), (311), (331). The difference in the two similar geometrical structures is because of the following:
1. Atomic form factors of K and Cl are similar, but of K and Br are very different
2. Atomic form factors of K and Cl are different, but of K and Br are similar
3. The structure factors of KCl and KBr are different
4. Structure factors of KCl and KBr are different and the form factors of K and Br are also similar.
Potassium has atomic number 19, while chlorine and bromine have atomic numbers 17 and 35, respectively. Because the $Cl^-$ and $K^+$ ions have the same number of electrons, their atomic form factors are nearly equal, and KCl "looks" to the X-rays as if it were a monatomic simple cubic lattice of lattice constant $a/2$. (A)
6. For any process, the second law of thermodynamics requires that the change of entropy of the universe be
1. Positive only
2. Positive or zero
3. Zero only
4. Negative or zero
(B) because for reversible process entropy is zero and for irreversible process entropy is positive

Problem set 2

1. What would be the approximate length of the day if the earth spun so fast that bodies floated on the equator? Take the radius of the earth $R=6\times 10^6 m$ and $g=9.8 m/s^2$
1. 12 hours
2. 6 hours
3. 3 hours
4. 1.5 hour
Body remains on earth's surface because
$gravitational~ pull ~(body~weight)\geq centrifugal~ force $
Body will start floating when
$centrifugal~force > gravitational~pull~i.e. body~weight$
$$m\omega^2r>mg$$ or $$\omega>\sqrt{\frac{g}{r}}$$ But $$\omega=\frac{2\pi}{T}$$ $$T(in~ seconds)>2\pi\sqrt{\frac{r}{g}}$$
$$T(in~ hours)>2\pi\sqrt{\frac{r}{g}}\frac{1}{3600}=\frac{2\times3.14\times1000\times\sqrt{6}}{3600\times\sqrt{9.8}}=1.5~ hours$$
2. The real matrix $A=\begin{pmatrix}a&-f&-g\\ f&a&h\\g&-h&a\end{pmatrix}$ is skew symmetric when
   1. $a=0$
   2. $f=0$
   3. $g=h$
   4. $f=g$ $
   Matrix $A$ is skew symmetric when $A_{ij}=-A_{ji}$. Hence, $a=0$
3. If $A$ and $B$ are matrices such that $AB=B$ and $BA=A$ then $A^2+B^2$ equals
1. 2AB
2. 2BA
3. A+B
4. AB
\begin{align*} A^2+B^2&=(BA)^2+(AB)^2\\ &=BABA+ABAB \end{align*} \begin{align*} A^2+B^2&=B\underline{AB}A+A\underline{BA}B\\ &=B\underline{BA}+A\underline{AB}\\ &=BA+AB=A+B \end{align*}
4. The average value of function $f(x)=4x^3$ in the interval 1 to 3 is
1. 15
2. 20
3. 30
4. 40
\begin{align*} <f(x)>&=\frac{\int_1^3f(x)dx}{\int_1^3dx}\\ &=\frac{\int_1^34^3dx}{\int_1^3dx}\\ &=\frac{[x^4]_1^3}{[x]_1^3}\\ &=\frac{81-1}{3-1}=40 \end{align*}

Problem set 1

Wave function of a particle moving in free space is given by, $\psi=e^{ikx}+2e^{-ikx}$. Find the energy of the particle.

one dimensional Schrödinger equation is $-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}+V\psi=E\psi$
For a free particle $V=0$
$-\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=E\psi$
$\frac{d^2\psi}{dx^2}=-k^2\left(e^{ikx}+2e^{-ikx}\right)=-k^2\psi$
$E=\frac{\hbar^2k^2}{2m}$