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

(A) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.Since $h$,$m$,and $

Vedclass · Accepted Answer

$\lambda / 2$

Vedclass · Answer

(A) The de-Broglie wavelength $\lambda$ of a particle of mass $m$ at temperature $T$ is given by $\lambda = \frac{h}{\sqrt{3mkT}}$.Since $h$,$m$,and $k$ are constants,we have $\lambda \propto \frac{1}{\sqrt{T}}$.Given $T_1 = 27^{\circ} C = 27 + 273 = 300 \ K$ and $T_2 = 927^{\circ} C = 927 + 273 = 1200 \ K$.Using the proportionality $\frac{\lambda_2}{\lambda_1} = \sqrt{\frac{T_1}{T_2}}$,we get $\frac{\lambda_2}{\lambda} = \sqrt{\frac{300}{1200}}$.$\frac{\lambda_2}{\lambda} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,$\lambda_2 = \frac{\lambda}{2}$.

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

Similar Questions

What is the de Broglie wavelength of an electron accelerated through a potential difference of $ 100 \ V $ (in $\text{Å}$)?

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 de Broglie wavelength $\lambda$ associated with a proton increases by $25 \%$,if its momentum is decreased by $p_0$. The initial momentum was

The de Broglie wavelength of a proton accelerated by a potential difference of $100 \ V$ is $\lambda_0$. If an alpha particle is accelerated by the same potential difference,its de Broglie wavelength will be:

$A$ proton is accelerated through a potential difference of $100 \ V$. To have the same de Broglie wavelength,what potential difference must be applied across a doubly ionized $_4^8 Be$ nucleus (in $V$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module