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Question: De Broglie wavelength of neutrons in thermal equilibrium is (given \({m_n} = 1.6 \times {10^{ - 27}}...

De Broglie wavelength of neutrons in thermal equilibrium is (given mn=1.6×10−27kg{m_n} = 1.6 \times {10^{ - 27}}kg)
(A)30.8TAo\left( A \right)\dfrac{{30.8}}{{\sqrt T }}{A^o}
(B)3.08TAo\left( B \right)\dfrac{{3.08}}{{\sqrt T }}{A^o}
(C)0.308TAo\left( C \right)\dfrac{{0.308}}{{\sqrt T }}{A^o}
(D)0.0308TAo\left( D \right)\dfrac{{0.0308}}{{\sqrt T }}{A^o}

Explanation

Solution

- Hint: In this question use the direct relation between De Broglie wavelength, mass of neutrons and temperature which is given as λ=h2mkT\lambda = \dfrac{h}{{\sqrt {2mkT} }} so use this direct relation to get the solution of the problem.

Formula used – λ=h2mkT\lambda = \dfrac{h}{{\sqrt {2mkT} }}

Complete step-by-step solution -
Thermal equilibrium
Thermal equilibrium is a state when two substances which are in contact exchange no heat with each other, i.e. in thermal equilibrium two substances are at equal temperature.
Consider the case when one hot body and one cold body make contact with other so the hot body transfer heat into the cold body till both the cold body and the hot body are at same temperature, this temperature is known as thermal equilibrium temperature often measured in degree Celsius or in Kelvin having symbol (oC)\left( {^oC} \right) or (K).
Given data:
Mass of neutrons = mn=1.6×10−27kg{m_n} = 1.6 \times {10^{ - 27}}kg
Now as we know that there is a direct relation between De Broglie wavelength, mass of neutrons and temperature which is given as
λ=h2mkT\lambda = \dfrac{h}{{\sqrt {2mkT} }}................... (1), where λ=\lambda = De Broglie wavelength.
h = Planck's constant = 6.626×10−346.626 \times {10^{ - 34}}
m = mass of neutrons
K = Boltzmann’s constant = 1.38×10−231.38 \times {10^{ - 23}}
T = temperature in degree Celsius.
Now substitute the given values in equation (1) we have,
⇒λ=6.626×10−342×(1.6×10−27)×(1.38×10−23)×T\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {2 \times \left( {1.6 \times {{10}^{ - 27}}} \right) \times \left( {1.38 \times {{10}^{ - 23}}} \right) \times T} }}
Now simplify the above equation we have,
⇒λ=6.626×10−344.416×10−50T\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{\sqrt {4.416 \times {{10}^{ - 50}}T} }}
⇒λ=6.626×10−342.1×10−25T\Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}}}{{2.1 \times {{10}^{ - 25}}\sqrt T }}
⇒λ=3.155×10−9T\Rightarrow \lambda = \dfrac{{3.155 \times {{10}^{ - 9}}}}{{\sqrt T }}
⇒λ=31.55×10−10Tm\Rightarrow \lambda = \dfrac{{31.55 \times {{10}^{ - 10}}}}{{\sqrt T }}m
⇒λ=31.55TAo\Rightarrow \lambda = \dfrac{{31.55}}{{\sqrt T }}{A^o}, [∵1A0=1×10−10m]\left[ {\because 1{A^0} = 1 \times {{10}^{ - 10}}m} \right]
⇒λ≃30.8TAo\Rightarrow \lambda \simeq \dfrac{{30.8}}{{\sqrt T }}{A^o}
So this is the required De Broglie wavelength.

Hence option (A) is the required answer.

Note: Whenever we face such types of questions always recall the relation between De Broglie wavelength, mass of neutrons and temperature which is stated above then just simply substitute the given value in this relation as above and simplify, we will get the required answer.