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PhysicsModern PhysicsMiscellaneousEasy2 minPYQ_2020
PhysicsEasysingle choice

De Broglie wavelength of0.05 eVthermal neutron is

Options:

Answer:
A
Solution:

λ=hp=h2m0K

λ=h2×1.66×10-27×0.05×1.6×10-19

λ=1.28 Å

λ≈1.3 Å

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:Modern PhysicsSubtopic:Miscellaneous
⏱ 2mℹ️ Source: PYQ_2020

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