Find the de-Broglie wavelength of neutron at 27°C. Given, Boltzmann constant, 1.38 x 10²³ J molecule^–1 k^–1, h = 6.63 x 10^–34 Js; mass of neutron 1.66 x 10^–23 kg.

Solution

Here,
Temperature, T = 27°C = 27 + 273 = 300 K.
Mass of neutron, m = 1.66 x 10^–23 kg.

Energy of neutron at temperature Tk is,

$E = \frac{3}{2} kT = \frac{3}{2} \times 1.38 \times 10^{- 23} \times 300 = 6.21 \times 10^{- 21} J$
Now,

$λ = \frac{h}{\sqrt{2 mE}} = \frac{6.63 \times 10^{- 34}}{\sqrt{2 \times 1.66 \times 10^{- 27} \times 6.21 \times 10^{- 21}}}$
i.e., $λ = 1.46 Å$ , is the de-broglie wavelength of the neutron.

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Find the de-Broglie wavelength of neutron at 27°C. Given, Boltzmann constant, 1.38 x 10²³ J molecule^–1 k^–1, h = 6.63 x 10^–34 Js; mass of neutron 1.66 x 10^–23 kg.

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Wave Optics

Find the de-Broglie wavelength of neutron at 27°C. Given, Boltzmann constant, 1.38 x 1023 J molecule–1 k–1, h = 6.63 x 10–34 Js; mass of neutron 1.66 x 10–23 kg.

Some More Questions From Wave Optics Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the de-Broglie wavelength of neutron at 27°C. Given, Boltzmann constant, 1.38 x 10²³ J molecule^–1 k^–1, h = 6.63 x 10^–34 Js; mass of neutron 1.66 x 10^–23 kg.