CBSE Class 12 Physics 2022 Term 2 Delhi Set 2 Solved Paper

Question : 3 of 4

Marks: +1, -0

An alpha particle is accelerated through a potential difference of

100 {\V}

. Calculate : 3
(i) The speed acquired by the alpha particle, and
(ii) The de-Broglie wavelength associated with it.
(Take mass of alpha particle

=6.4 × 10^{-27} {\kg}

)

Solution:

1/2 mv^2 = qV

or,

1/2 mv^2 = 2e×100

or,

mv^2 = 400 eV

or,

v=√{{400eV}/m}

or,

v=√{{400×1.6×10^{-19}}/{6.4 × 10^{-27}}}

∴ v=10^5 m\/s

(ii) de-Brogile wavelength

=λ=h/√2mqV

or,

λ= {6.6×10{-34}}/√{2×6.4×10^{-27}×2×1.6×10^{-19}×100}

∴ λ=1.03 × 10^{-12}m

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