If the mass of neutron is $$1.7x$$ $$10-27$$ kg, then the $$de-Broglie$$ wavelength of neutron of energy $$3$$ eV is: (h=$$ $$6.6x10-34$$ $$Js)

Question

If the mass of neutron is $$1.7x$$ $$10-27$$ kg, then the $$de-Broglie$$ wavelength of neutron of energy $$3$$ eV is: (h=$$ $$6.6x10-34$$ $$Js)

Infinity Learn · Accepted Answer

mn $$=$$  $$1.7$$  ×  $$10$$−$$27$$  kg               $$E=3eV$$,      λ$$=$$?$$E=P22m$$   ⇒   $$P=2mE$$λ  $$=$$  $$h2mE$$  $$=$$  $$6.6$$ ×  $$10$$−$$342$$ ×  $$1.7$$  ×  $$10$$−$$27$$  × $$3$$  ×  $$1.6$$  ×  $$10$$−$$19$$                       $$=$$  $$6.6$$  ×  $$10$$−$$344.04$$  ×  $$10$$−$$23$$                       $$=1.65$$  ×  $$10$$−$$11$$  m

Matter waves (DC Broglie waves)

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If the mass of neutron is $1.7 x 10^{- 27} k g$ , then the de-Broglie wavelength of neutron of energy 3 eV is: $(h = 6.6 x 10^{- 34} Js)$

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Matter waves (DC Broglie waves)

Remember concepts with our Masterclasses.

If the mass of neutron is 1.7x 10-27 kg, then the de-Broglie wavelength of neutron of energy 3 eV is: (h= 6.6x10-34 Js)

mn = 1.7 × 10−27 kg E=3eV, λ=?E=P22m ⇒ P=2mEλ = h2mE = 6.6 × 10−342 × 1.7 × 10−27 × 3 × 1.6 × 10−19 = 6.6 × 10−344.04 × 10−23 =1.65 × 10−11 m

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free study material

If the mass of neutron is $1.7 x 10^{- 27} k g$ , then the de-Broglie wavelength of neutron of energy 3 eV is: $(h = 6.6 x 10^{- 34} Js)$