Test Index

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

© examsnet.com
Question : 3 of 4
Marks: +1, -0
An alpha particle is accelerated through a potential difference of 100V100 \text{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 =6.4×10−27kg=6.4 \times 10^{-27} \text{kg} )
Solution:  
12mv2=qV\frac{1}{2} m v^2 = q V
or, 12mv2=2e×100\frac{1}{2} m v^2 = 2e \times 100
or, mv2=400eVm v^2 = 400 \text{eV}
or, v=400eVmv = \sqrt{\frac{400 \text{eV}}{m}}
or, v=400×1.6×10−196.4×10−27v = \sqrt{\frac{400 \times 1.6 \times 10^{-19}}{6.4 \times 10^{-27}}}
∴v=105m/s\therefore v = 10^5 \text{m/s}
(ii) de-Brogile wavelength
=λ=h2mqV= \lambda = \frac{h}{\sqrt{2 m q V}}
or, λ=6.6×10−342×6.4×10−27×2×1.6×10−19×100\lambda = \frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6.4 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times 100}}
∴λ=1.03×10−12m\therefore \lambda = 1.03 \times 10^{-12} \text{m}
© examsnet.com
Go to Question:
Prev Question
Next Question