The de Broglie wavelength associated with a n | vedclass

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

Question

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.For a neutron in thermal equilibrium at temperature $T$,the kinetic energy is $E = \

Vedclass · Accepted Answer

$\frac{30.7}{\sqrt{T}} \ \mathring{A}$

Vedclass · Answer

(C) The de Broglie wavelength is given by $\lambda = \frac{h}{p}$.For a neutron in thermal equilibrium at temperature $T$,the kinetic energy is $E = \frac{p^2}{2m_n} = \frac{3}{2}kT$. However,using the approximation $E \approx kT$ as per the question:$p = \sqrt{2m_n E} = \sqrt{2m_n kT}$.Substituting the values: $h = 6.63 	imes 10^{-34} \ J\cdot s$,$m_n = 1.675 	imes 10^{-27} \ kg$,and $k = 1.38 	imes 10^{-23} \ J/K$.$\lambda = \frac{h}{\sqrt{2m_n kT}} = \frac{6.63 	imes 10^{-34}}{\sqrt{2 	imes 1.675 	imes 10^{-27} 	imes 1.38 	imes 10^{-23} 	imes T}}$.$\lambda = \frac{6.63 	imes 10^{-34}}{\sqrt{4.623 	imes 10^{-50} 	imes T}} = \frac{6.63 	imes 10^{-34}}{2.15 	imes 10^{-25} \sqrt{T}}$.$\lambda \approx \frac{3.08 	imes 10^{-9}}{\sqrt{T}} \ m = \frac{30.8}{\sqrt{T}} \ \mathring{A}$.Rounding to the nearest standard value,we get $\lambda = \frac{30.7}{\sqrt{T}} \ \mathring{A}$.

The de Broglie wavelength associated with a neutron at temperature $T$ is given by: $(E = kT)$

Similar Questions

The de Broglie wavelength of a proton and $\alpha$-particle are equal. The ratio of their velocities is ...... .

The kinetic energy of an electron gets tripled,then the de-Broglie wavelength associated with it changes by a factor of:

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy,$E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

When energy is added to an electron,its de Broglie wavelength decreases from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$. The added energy is:

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of the proton,$\lambda_2$ is the wavelength associated with the photon,and the energy of the photon is $E$,then $(\lambda_1 / \lambda_2)$ is proportional to

Vedclass Test Series

Exam Paper Generator

Online Exam Module