If the de Broglie wavelength of a neutron at | vedclass

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(A) The de Broglie wavelength $\lambda$ of a neutron at temperature $T$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mKT}}$.From this,we can se

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$\frac{\lambda}{2}$

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(A) The de Broglie wavelength $\lambda$ of a neutron at temperature $T$ is given by the formula: $\lambda = \frac{h}{\sqrt{3mKT}}$.From this,we can see that $\lambda \propto \frac{1}{\sqrt{T}}$.Given initial temperature $T_i = 77^{\circ}C = 77 + 273 = 350 \ K$.Given final temperature $T_f = 1127^{\circ}C = 1127 + 273 = 1400 \ K$.Using the proportionality $\frac{\lambda_f}{\lambda_i} = \sqrt{\frac{T_i}{T_f}}$,we get:$\frac{\lambda_f}{\lambda} = \sqrt{\frac{350}{1400}} = \sqrt{\frac{1}{4}} = \frac{1}{2}$.Therefore,the final wavelength is $\lambda_f = \frac{\lambda}{2}$.

If the de Broglie wavelength of a neutron at a temperature of $77^{\circ} C$ is $\lambda$,then the de Broglie wavelength of the neutron at a temperature of $1127^{\circ} C$ is

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