You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

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 $x$ ?

Open in App

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 $= \frac{1}{2} m v^{2}$
$= 2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{6} = 6.4 \times 10^{- 13}$

Suggest Corrections

Similar questions

Q. An alpha particle after passing through a potential difference of

2 \times 10^{6} v o l t

falls on a silver foil. The atomic number of silver is

47

. Calculate

(I)

the

K . E

of the alpha-particle at the time of falling on the foil.

(I I) K . E

of the alpha-particle at a distance of

5 \times 10^{- 14} m

from the nucleus,

(I I I)

the shortest distance from the nucleus of silver to which the alpha-particle reaches.

Q. An alpha particle after passing through a potential difference of

2 \times 10^{6}

volt falls on a silver foil. The atomic number of silver is 47. The K.E. of the

α

-particle at the time of falling on the foil is:

In Rutherford's alpha particle scattering experiment, when fast moving alpha particles were made to fall on a thin gold foil then most of the alpha particles passed straight through the foil. Which of the given conclusion is correct?

Q. When fast moving alpha particles are made to fall on a thin gold foil, most of them go straight through the foil because:

Q. When a silver foil (Z = 47) was used in a

α

-ray scattering experiment, the number of

α

particles scattered at

30^{\circ}

was found to be 200 per minute. If the silver foil is replaced by aluminium (z = 13) foil of same thickness, the number of

α

-particles scattered per minute at

30^{\circ}

is nearly equal to

Join BYJU'S Learning Program

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 $= \frac{1}{2} m v^{2}$
$= 2 \times 1.6 \times 10^{- 19} \times 2 \times 10^{6} = 6.4 \times 10^{- 13}$