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Question: The de-Broglie wavelength of a neutron at \[{27^o}C\] is \(\lambda \) . What will be its wavelength ...

The de-Broglie wavelength of a neutron at 27oC{27^o}C is λ\lambda . What will be its wavelength at 927oC{927^o}C?
A. λ2\dfrac{\lambda }{2}
B. λ3\dfrac{\lambda }{3}
C. λ4\dfrac{\lambda }{4}
D. λ9\dfrac{\lambda }{9}

Explanation

Solution

Use the de Broglie wavelength equation. Express the momentum in terms of kinetic energy. Substitute kinetic energy with the equation of energy of thermal neutrons.

Complete step by step answer:
Let T1=27oC=300K{T_1} = {27^o}C = 300K and T2=927oC=1200K{T_2} = {927^o}C = 1200K
Then the corresponding de Broglie wavelength will be λ1{\lambda _1} and λ2{\lambda _2}
Thus for the first case
λ1=h3mkT1{\lambda _1} = \dfrac{h}{{\sqrt {3mk{T_1}} }}
Then for the second case
λ2=h3mkT2{\lambda _2} = \dfrac{h}{{\sqrt {3mk{T_2}} }}
We divide both equations and cancel the common terms.
After that we get the following ratio
λ1λ2=T2T1\dfrac{{{\lambda _1}}}{{{\lambda _2}}} = \sqrt {\dfrac{{{T_2}}}{{{T_1}}}}
Thus the required wavelength will be
λ2=T1T2λ=3001200λ=λ2{\lambda _2} = \sqrt {\dfrac{{{T_1}}}{{{T_2}}}} \lambda = \sqrt {\dfrac{{300}}{{1200}}} \lambda = \dfrac{\lambda }{2}

So, the correct answer is “Option A”.

Additional Information:
De Broglie wavelength is expressed as Planck's constant divided by momentum. It states that every particle in this universe removes in a wave like pattern. However for some particles the speed is very less. Thus the momentum is less and thus the wavelength is very large. The wavelength is so large that the object is seen as moving in a straight line unlike that of a wave like manner. However for objects moving with a very high speed, such a speed that is comparable to the speed of light, the momentum increases and the wavelength decreases. Hence we can clearly see the particles moving in a wave like pattern. Normally larger objects move with a larger wavelength and small objects move with a smaller and evident wavelength.

Note:
Students should learn the different ways the kinetic energy can be expressed. Some questions demand other expressions for momentum as well as kinetic energy. Also do carefully cancel the terms.
The wavelength is so large that the object is seen as moving in a straight line unlike that of a wave like manner.