-12 An electron microscope operates with electrons of wavelength 2.74 pm (2.74 x 10¯ m). Find the kinetic energy these electrons.

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HW5 Q10

An electron microscope operates with electrons of wavelength 2.74 pm (2.74 × 10-¹² m).
Find the kinetic energy these electrons.
201 keV
-14
6.43 107 eV
-14
3.22 × 107 eV
400 keV
Transcribed Image Text:An electron microscope operates with electrons of wavelength 2.74 pm (2.74 × 10-¹² m). Find the kinetic energy these electrons. 201 keV -14 6.43 107 eV -14 3.22 × 107 eV 400 keV
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i dont understand this (Kinetic energy of electron isKE=12mv2        =12m2v2mKE=p22m..................(1)AlsoDe Broglie wavelength of electron is λ=hmvλ=hpp=hλBy equation (1) we getKE=p22m       =(hλ)22m     =12m×(hλ)2    =19.1×10-31×(6.626×10-342.74×10-12)2   =3.213145×10 -14​​​​   J   =3.213145×10 -141.6×10-19  eV   =200821  eV   =201  keV) 

Please have it written in a legible way so i can understand how to get 201 keV

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