KVPY SA Exam 19-Nov-2017 Question Paper

Question : 41 of 80

Marks: +1, -0

C_60

emerging from a source at a speed (v) has a de Broglie wavelength of 11.0 Å. The value of v (in m s

^{–1}

) is closest to
[Planck’s constant h = 6.626 × 10

^{–34}

J s]

Solution:

λ=h/{mv}⇒v=h/{m.λ}

={6.62 ×10^{-34}}/{720× 10^{-3} ×11 ×10^{-10}}

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