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MODEL PAPER 9 — Physics STD 12 Science — Question

Bihar BoardEnglish MediumSTD 12 SciencePhysicsMODEL PAPER 93 Marks
Question
An alpha particle is accelerated through a potential difference of 100 V. Calculate:
i. The speed acquired by the alpha particle, and
ii. The de-Broglie wavelength associated with it. 
(Take mass of alpha particle $=6.4 \times 10^{-27} kg$ )
✓

Answer

i. $\frac{1}{2} m v^2=q V$
\frac{1}{2} m v^2=2 e \times 100 \\ 
\text { or, } \frac{1}{2} m v^2=2 e \times 100 \\ 
\text { or, } m v^2=400 eV \end{array}$
$\begin{array}{l}\text { or, } v =\sqrt{\frac{400 eV }{m}} \\ 
\text { or, } v =\sqrt{\frac{400 \times 1.6 \times 10^{-19}}{6.4 \times 10^{-27}}} \\ \therefore v =10^5 m / s \end{array}$
ii. de-Broglie wavelength $=\lambda=\frac{h}{\sqrt{2 m q V}}$
$\begin{array}{l}\text { Or, } \lambda=\frac{6.6 \times 10^{-34}}{\sqrt{2 \times 6.4 \times 10^{-27} \times 2 \times 1.6 \times 10^{-19} \times 100}} \\ 
\therefore \lambda=1.03 \times 10^{-12} m\end{array}$

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