Fringe Position in Young's Double Slit Experiment
The position ($y$) of the $k^{th}$ bright fringe in a double-slit experiment is given by the formula:
$ y_k = \frac{k \lambda L}{d} $
where $k$ is the fringe order, $\lambda$ is the wavelength of the particles, $L$ is the distance to the screen, and $d$ is the slit separation.
De Broglie Wavelength and Kinetic Energy
The de Broglie wavelength ($\lambda$) is related to momentum ($p$) by $\lambda = \frac{h}{p}$, where $h$ is Planck's constant.
Kinetic energy ($KE$) is related to momentum ($p$) and mass ($m$) by $KE = \frac{p^2}{2m}$.
Since the kinetic energies of both $C_{60}$ and $C_{70}$ beams are the same ($KE_{60} = KE_{70}$):
$ \frac{p_{60}^2}{2m_{60}} = \frac{p_{70}^2}{2m_{70}} $
This implies:
$ \frac{p_{70}}{p_{60}} = \sqrt{\frac{m_{60}}{m_{70}}} $
The mass of a fullerene molecule is proportional to the number of carbon atoms. Thus, $m_{60} \propto 60$ and $m_{70} \propto 70$.
$ \frac{p_{70}}{p_{60}} = \sqrt{\frac{60}{70}} $
Since $\lambda = h/p$, the ratio of wavelengths is inversely proportional to the ratio of momenta:
$ \frac{\lambda_{60}}{\lambda_{70}} = \frac{p_{70}}{p_{60}} = \sqrt{\frac{60}{70}} $
Calculating the Corresponding Fringe Order (n)
We are given that the position of the 4th bright fringe for $C_{60}$ matches the position of the $n^{th}$ bright fringe for $C_{70}$.
$ y_{4, C_{60}} = y_{n, C_{70}} $
Using the fringe position formula:
$ \frac{4 \lambda_{60} L}{d_{60}} = \frac{n \lambda_{70} L}{d_{70}} $
The distance to the screen ($L$) cancels out:
$ \frac{4 \lambda_{60}}{d_{60}} = \frac{n \lambda_{70}}{d_{70}} $
Solving for $n$:
$ n = 4 \times \frac{\lambda_{60}}{\lambda_{70}} \times \frac{d_{70}}{d_{60}} $
Substitute the known values and the wavelength ratio:
- $d_{60} = 50$ nm
- $d_{70} = 92.5$ nm
- $\frac{\lambda_{60}}{\lambda_{70}} = \sqrt{\frac{60}{70}}$
$ n = 4 \times \sqrt{\frac{60}{70}} \times \frac{92.5 \text{ nm}}{50 \text{ nm}} $
Calculate the value of $n$:
$ n = 4 \times \sqrt{1.1667} \times 1.85 $
$ n \approx 4 \times 1.0801 \times 1.85 $
$ n \approx 4 \times 1.9982 \approx 7.9928 $
Rounding to the nearest integer, $n = 8$.
Important Questions from Schrödinger Equation 1D Potentials Harmonic Oscillator
- The energy $E$ and degeneracy $d$ of the second excited state of a three-dimensional, isotropic quantum harmonic oscillator with angular frequency $\omega$ are
- A particle of mass $ m $ is in a potential $ V(x) = \frac{1}{2}m\omega^2x^2 $ for $ x > 0 $ and $ V(x) = \infty $ for $ x \leq 0 $, where $ \omega $ is the angular frequency. The ratio of the energies corresponding to the lowest energy level to the next higher level is
- Consider a particle in a two dimensional infinite square well potential of side $L$, with $0 \le x \le L$ and $0 \le y \le L$. The wavefunction of the particle is zero only along the line $y = \frac{L}{2}$, apart from the boundaries of the well. If the energy of the particle in this state is $E$, what is the energy of the ground state?
The wavefunction of a particle in an infinite one-dimensional potential well at time $t$ is
$\Psi(x, t) = \sqrt{\frac{2}{3}} e^{-iE_1t/\hbar}\psi_1(x) + \frac{1}{\sqrt{6}} e^{i\pi/6}e^{-iE_2t/\hbar}\psi_2(x) + \frac{1}{\sqrt{6}} e^{i\pi/4}e^{-iE_3t/\hbar}\psi_3(x)$
where $\psi_1, \psi_2$ and $\psi_3$ are the normalized ground state, the normalized first excited state and the normalized second excited state, respectively. $E_1, E_2$ and $E_3$ are the eigen-energies corresponding to $\psi_1, \psi_2$ and $\psi_3$, respectively. The expectation value of energy of the particle in state $\Psi(x, t)$ isA particle is subjected to a potential
$V(x) = \begin{cases} \infty, & x \le 0 \\ V_0, & a \le x \le b \\ 0, & \text{elsewhere} \end{cases}$
Here, $a > 0$ and $b > a$. If the energy of the particle $E < V_0$, which one of the following schematics is a valid quantum mechanical wavefunction ($\Psi$) for the system?
