CBSE Class 12 · 2022 · Modern PhysicsShort Answer

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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 1 0^{- 27} kg

Answer: $v = 1 0^{5} m/s$ ; $λ \approx 1.0 \times 1 0^{- 12} m$ .

Solution

Charge of

α

-particle

q = 2 e = 3.2 \times 1 0^{- 19} C

.
(i) From

q V = \frac{1}{2} m v^{2}

v = \frac{2 q V}{m} = \frac{2 ( 3.2 \times 1 0 ^{- 19} ) ( 100 )}{6.4 \times 1 0 ^{- 27}} = 1 0^{10} = 1.0 \times 1 0^{5} m/s .

(ii) de-Broglie wavelength:

λ = \frac{h}{m v} = \frac{6.63 \times 1 0 ^{- 34}}{( 6.4 \times 1 0 ^{- 27} ) ( 1.0 \times 1 0 ^{5} )} \approx 1.0 \times 1 0^{- 12} m .

v = 1.0 \times 1 0^{5} m/s, λ \approx 1.0 \times 1 0^{- 12} m

CBSE Class 12 Board · 2022 · Subjective

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