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

Question : 3

Total: 4

An alpha particle is accelerated through a potential difference of 100V. Calculate : 3
(i) The speed acquired by the alpha particle, and
(ii) The de-Broglie wavelength associated with it.
(Take mass of alpha particle =6.4×10−27‌kg )

Solution:

mv2=qV
or,

mv2=2e×100
or, mv2=400eV
or, v=√

400eV

or, v=√

400×1.6×10−19

6.4×10−27

∴v=105m∕s
(ii) de-Brogile wavelength
=λ=

√2

mqV
or, λ=

6.6×10−34

√2×6.4×10−27×2×1.6×10−19×100

∴λ=1.03×10−12m

Go to Question:

Prev Question

Next Question