Question

An alpha particle after passing through a potential difference

Answer 1

$\huge\mathbb\red{BONJOUR,}$

$\huge\mathfrak{Answer:}$

An alpha particle after passing through a potential difference of 2 * 10^6. An alpha particle after passing through a potential difference of 2 * 10^6 volt falls on a silver foil. The atomic number of silver is 47. ... 3) The shortest distance from the nucleus of silver to which the alpha-particle reaches.

$<font color ="purple"><marquee behavior ="alternate"> ~♥~~♥~Hope It Helps~♥~~♥~</font color ="purple "></marquee behavior="alternate">$

$\huge\mathfrak{Regards}$

$\huge\mathbb\red{ITZAARAV}$

Answer 2

Answer:An alpha particle after passing through a potential difference of 2 * 10^6 volt falls on a silver foil. The atomic number of silver is 47. Calculate: 1) The K.E. of the alpha-particle at the time of falling on the foil. 2) K.E. of the alpha-particle at a distance of 5 * 10^-14 m from nucleus. 3) The shortest distance from the nucleus of silver to which the alpha-particle reaches.

answer;Given  

Potential difference = 2×106

Atomic number of silver = 47

Charge of alpha particle=2e

Charge of silver = 47e(1)  

Mass of alpha particle = 12mv2=2×1.6×10−19×2×106=6.4×10−13

K.E at A = K.E at B + P.E AT B  

6.4×10−13 = K.E AT B +  P.E at B

K.E at B = 2.1×10−13 joules

Explanation: