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Monday, 7 November 2016
Problem set 21
Four charges (two +q and two -q) are kept fixed at the four vertices of a square of side a as shown
At the point P which is at a distance R from the centre ( R > > a), the potential is proportional to
1/R
1/R^2
1/R^3
1/R^4
The system forms a quadrupole. The electric field and potential energy variations for various system are as follow:
For monopole E\propto 1/r^2 and V\propto 1/r
For dipole E\propto 1/r^3 and V\propto 1/r^2
For quadrupole E\propto 1/r^4 and V\propto 1/r^3
For octapole E\propto 1/r^5 and V\propto 1/r^4
Hence, answer is (C).
A point charge q of mass m is kept at a distance d below a grounded infinite conducting sheet which lies in the xy-plane. For what value of d will the charge remains stationary?
q/4\sqrt{mg\pi\epsilon_0}
q/\sqrt{mg\pi\epsilon_0}
There is no finite value of d
\sqrt{mg\pi\epsilon_0}/q
When a charge +q is placed below the plate at a distance d, then there is an image charge -q above the plate at the same distance d. For the charge to remain stationary, weight mg must equal to electrostatic force \frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}. i.e.
mg=\frac{1}{4\pi\epsilon_0}\frac{q^2}{(2d)^2}d=\frac{q}{4\sqrt{mg\pi\epsilon_0}}
Hence, answer is (A).
The wave function of a state of the hydrogen atom is given by {\scriptstyle\Psi=\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}}, where \psi_{nlm} is the normalized eigenfunction of the state with quantum numbers n,l and m in the usual notation. The expectation value of L_z in the state \Psi is
15\hbar/16
11\hbar/16
3\hbar/8
\hbar/8
Let the normalized wavefunction is given by
\Psi={\scriptstyle A\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}
Applying normalization condition
\begin{align*}
|\Psi|^2&=|A|^2\left[|\psi_{200}|^2+|2\psi_{211}|^2\right.\\
&\left.+|3\psi_{210}|^2+|\sqrt{2}\psi_{21-1}|^2\right]\\
&=1
\end{align*}
we get, |A|^2\left[1+4+9+2\right]=1, hence, A=\frac{1}{4}. Hence, normalized wavefunction is
\Psi={\scriptstyle \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]}\begin{align*}
&< L_z > =\left < \Psi|L_z|\Psi\right > \\
&=\left < \frac{1}{4}\left[\psi_{200}+2\psi_{211}+3\psi_{210}\right.\right.\\
&\left.+\sqrt{2}\psi_{21-1}\right]\left|L_z\right|\frac{1}{4}\left[\psi_{200}\right.\\
&\left.\left.+2\psi_{211}+3\psi_{210}+\sqrt{2}\psi_{21-1}\right]\right >\\
&=\frac{1}{16}\left\{\left < \psi_{200}\left|L_z\right|\psi_{200}\right >\right.\\
&+4\left < \psi_{211}\left|L_z\right|\psi_{211}\right >+9\left < \psi_{210}\left|L_z\right|\psi_{210}\right >\\
&\left.+2\left < \psi_{21-1}\left|L_z\right|\psi_{21-1}\right >\right\}
\end{align*}
But L_z|\psi_{nlm}> =m\hbar|\psi_{nlm}>\begin{align*}
< L_z >&=\frac{1}{16}\left\{0+4\hbar+0+2(-1)\hbar\right\}\\
&=\frac{1}{8}\hbar
\end{align*}
Hence, answer is (D)
The energy eigenvalues of a particle in the potential V(x) =\frac{1}{2}m\omega^2x^2-ax are
\begin{align*}
V(x)&=\frac{1}{2}m\omega^2x^2-ax=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}\right]\\
&=\frac{1}{2}m\omega^2\left[x^2-\frac{2ax}{m\omega^2}+\frac{a^2}{m^2\omega^4}-\frac{a^2}{m^2\omega^4}\right]\\
&=\frac{1}{2}m\omega^2\left[\left(x-\frac{a}{m\omega^2}\right)^2-\left(\frac{a}{m\omega^2}\right)^2\right]
\end{align*}
Let \left(x-\frac{a}{m\omega^2}\right)=x'\begin{align*}
V(x')&=\frac{1}{2}m\omega^2x'^2-\frac{1}{2}m\omega^2\left(\frac{a}{m\omega^2}\right)^2\\
&=\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}
\end{align*}
Second term on RHS is a constant.
H'=-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}\begin{align*}
E&=\left < \psi_n|H'|\psi_n\right > \\
&=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2-\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\
&=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\left < \psi_n\left|\frac{a^2}{2m\omega^2}\right|\psi_n\right >\\
&=\left < \psi_n\left|-\frac{\hbar^2}{2m}\frac{d^2}{dx'^2}+\frac{1}{2}m\omega^2x'^2\right|\psi_n\right >-\frac{a^2}{2m\omega^2}\left < \psi_n\left.\right|\psi_n\right >\\
&=\left(n+\frac{1}{2}\right)\hbar\omega-\frac{a^2}{2m\omega^2}
\end{align*}
Hence, answer is (A).
If a particle is represented by the normalized wave function
\begin{align*}
\psi(x)=\begin{cases}\frac{\sqrt{15}\left(a^2-x^2\right)}{4a^{5/2}}&{\scriptstyle\text{for} -a < x < a}\\
0 &\text{otherwise}
\end{cases}
\end{align*}
the uncertainty \Delta p in its momentum is
2\hbar/5a
5\hbar/2a
\sqrt{10}\hbar/a
\sqrt{5}\hbar/\sqrt{2}a
\Delta P=\sqrt{<\Delta P^2>-<\Delta P>^2}\begin{align*}
& <\Delta P>=\int\psi^*\left(-i\hbar\frac{\partial}{\partial x}\right)\psi\:dx\\
&=-\frac{15i\hbar}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial(a^2-x^2)}{\partial x}dx\\
&=\frac{15i\hbar}{a^58}\int_{-a}^{a}(a^2-x^2)x\:dx=0
\end{align*}
Since the integrand is an odd function.
\begin{align*}
& <\Delta P^2>=\int\psi^*\left(-\hbar^2\frac{\partial^2}{\partial x^2}\right)\psi\:dx\\
&=-\frac{15\hbar^2}{a^516}\int\limits_{-a}^{a}(a^2-x^2)\frac{\partial^2(a^2-x^2)}{\partial x^2}dx\\
&=\frac{15\hbar^2}{a^58}\int_{-a}^{a}(a^2-x^2)\:dx\\
&=\frac{15\hbar^2}{a^58}\left[a^2x-\frac{x^3}{3}\right]_{-a}^{a}=\frac{5\hbar^2}{2a^2}
\end{align*}\Delta P=\frac{\sqrt{5}\hbar}{\sqrt{2}a}
Hence, answer is (D).
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