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Standard XII

Chemistry

1900's Physics and Wave

The K.E. of t...

Question

The K.E. of the alpha-particle at the time of falling on the foil is $x \times 10^{- 13}$ . What is $x$ ?

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Solution

Potential difference $= 2 \times 10^{6}$
Atomic number of silver $= 47$
Charge on alpha-particle $= 2 e^{-}$
Charge on silver $= 47 e (1)$
Hence the mass of particle $= \frac{1}{2} m v^{2}$
$= 2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{6} = 6.4 \times 10^{- 13}$

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The K.E. of the alpha-particle at the time of falling on the foil is x×10−13. What is x?

Potential difference =2×106Atomic number of silver =47Charge on alpha-particle =2e−Charge on silver =47e(1)Hence the mass of particle =12mv2=2×1.6×10−19×2×106=6.4×10−13

The K.E. of the alpha-particle at the time of falling on the foil is $x \times 10^{- 13}$ . What is $x$ ?

Potential difference $= 2 \times 10^{6}$
Atomic number of silver $= 47$
Charge on alpha-particle $= 2 e^{-}$
Charge on silver $= 47 e (1)$
Hence the mass of particle $= \frac{1}{2} m v^{2}$
$= 2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{6} = 6.4 \times 10^{- 13}$