Q.

An electron is moving with a kinetic energy of  4.55×1025J. Its de-Broglie wavelength (in nm) is

me=9.1×1031kg,h=6.63×1034Js

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a

728.5

b

72.85

c

628.5

d

7285

answer is B.

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Detailed Solution

λ=h2meKEe

λ=6.63×10-342×9.1×10-31×4.55×10-25=6.63×10-349.1×10-28

λ=6.63×10-349.1×10-28=0.728×10-6 m

λ=0.728×10-6=0.7285×109 nm =728.5 nm

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