The de-Broglie wavelength of a neutron at 27^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The de-Broglie wavelength $\lambda$ of a neutron is given by $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $E_k$ is the kinetic energy.For a neutron in

Vedclass · Accepted Answer

$\frac{\lambda_0}{2}$

Vedclass · Answer

(C) The de-Broglie wavelength $\lambda$ of a neutron is given by $\lambda = \frac{h}{\sqrt{2mE_k}}$,where $E_k$ is the kinetic energy.For a neutron in thermal equilibrium at temperature $T$,the average kinetic energy is $E_k = \frac{3}{2} k_B T$.Thus,$\lambda = \frac{h}{\sqrt{2m(\frac{3}{2} k_B T)}} = \frac{h}{\sqrt{3mk_B T}}$.This implies $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.Therefore,$\frac{\lambda_2}{\lambda_0} = \sqrt{\frac{T_1}{T_2}} = \sqrt{\frac{300}{1200}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.So,$\lambda_2 = \frac{\lambda_0}{2}$.

The de-Broglie wavelength of a neutron at $27^{\circ} C$ is $\lambda_0$. What will be its wavelength at $927^{\circ} C$?

Similar Questions

If the momentum of an electron is changed by $\Delta p,$ then the de-Broglie wavelength associated with it changes by $0.50\%.$ The initial momentum of the electron will be:

The ratio of de Broglie wavelengths associated with thermal neutrons at temperatures $127^{\circ} C$ and $352^{\circ} C$ is

An electron (mass $m$) with initial velocity $\overrightarrow{v} = v_{0} \hat{i} + v_{0} \hat{j}$ is in an electric field $\overrightarrow{E} = -E_{0} \hat{k}$. If $\lambda_{0}$ is the initial de-Broglie wavelength of the electron,its de-Broglie wavelength at time $t$ is given by:

If the momentum of an electron is changed by $\Delta P,$ then the de Broglie wavelength associated with it changes by $0.5\%.$ The initial momentum of the electron will be

If an electron and a photon propagate in the form of waves having the same wavelength,it implies that they have the same

Vedclass Test Series

Exam Paper Generator

Online Exam Module