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Question

The de-Broglie wavelength of a neutron at 927 degrees Celsius is lambda. what will be its wavelength at 27-degree Celsius?


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Solution

Step 1:Given Data

de-Broglie wavelength at 927οc, λ1=λ.

Temperature, T1=927+273K=1200K

Temperature, T2=273+27K=300K

Step 2:Relation between de-Broglie wavelength and temperature is given below:

λneutron=CT

Step 3:Substitute 1200K for T and λ for λneutron in above relation:

λ=C1200K (1)

Step 4:Substitute 300K for T1 and λ2 for λneutron in above relation:

λ2=C300K (2)

Step 5:Divide equation 1 in equation 2.

λ2λ=1200K300Kλ2λ=2λ2=2λ

Hence, the Wavelength at 27-degree Celsius is 2λ.


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