De-broglie wavelength of a neutron at 927 degree Celsius is lambda. what will be its wavelength at 27-degree Celsius?

Answer:

[latex]\lambda _{neutron}\alpha \frac{1}{\sqrt{T}}[/latex]

⇒ [latex]\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{T_{2}}{T_{1}}}[/latex]

⇒ [latex]\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{273+927}{273+27}}[/latex]

⇒ [latex]\frac{\lambda _{1}}{\lambda _{2}} = \sqrt{\frac{1200}{300}}[/latex]

⇒ [latex]\frac{\lambda _{1}}{\lambda _{2}} = 2[/latex]

λ2 = λ/2

 

 

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