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Q.

The kinetic energy of an electron with de-Broglie wavelength of 0.3 nanometre is

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a

0.168 eV

b

16.8 eV

c

1.68 eV

d

2.5 eV

answer is B.

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Detailed Solution

λ=h2mE⇒E=h22mλ2   =(6.6×10-34)22×9.1×10-31×(0.3×10-9)2   =2.65×10-18J=16.8 eV
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