Question

an electron beam emerged from an accelerator with kinetic energy of 100ev.what is its de-broglie wavelength?

Answer 1

Answer:

de broglie wavelength=1.23×10^-10 m/1.23A°

Explanation:

Given:K.E=100ev

=1.6×10^-17J[1ev=1.6×10^-19J]

de broglie wavelength=h/mv

• =6.67×10^-34/p[p=mv]
• =6.67 ×10^-34/√2mK.E [square root of 2m K.E]
• =6.67×10^-34/√2×9.1×10^-31×1.6×10^-17[square root of 2×9.1×10^-31×1.6×10^-17]
• =6.67 ×10^-33/5.40 ×10-24
• =1.23 ×10^-10
• =1.23A°

Answer 2

Explanation:

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