An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron

Question

An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron

An electron with initial kinetic energy of `100 eV` is acclerated through a potential difference of `50 V`. Now the de-Brogile wavelength of electron becomes
A. `1 Å`
B. `sqrt(1.5Å)`
C. `sqrt(3)Å`
D. `12.27Å`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

Salim Ahmed Kawsar

Correct Answer - A
For electron `lambda_(db)=sqrt(150/(100+50))=1 Å`

4 views Answered 2023-02-05 15:29