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  4. The de-Broglie wavelength of a neutron at 927° C is λ. What will be its wavelength at 27° C ?

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

Bihar CECEBihar CECE 2009Dual Nature of Radiation and Matter

Solution:

de-Broglie wavelength $\lambda \propto \frac{1}{\sqrt{T}}$
$\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{T_{2}}{T_{1}}}$
$\frac{\lambda_{1}}{\lambda_{2}} =\sqrt{\frac{273+927}{273+27}}$
$=\sqrt{\frac{1200}{300}}=2$
or $\lambda_{2} =\frac{\lambda}{2}$