I need a pic so I can give you an answer
Answer: 1.3 ×10^-31
Explanation:
the required probability is P = e^(-2αL)
Firstly, evaluate (-2αL)
α= 1/hc √2mc^2 (U - E)
h= modified planck's constant
where,
(-2αL)
= -(2L)/(h/2π ) ×√2mc^2 (U - E)
= -(2L) / (hc^2/π )×√2mc^2 (U - E)
(hc^2/2pi) = 197*eV.nm (standard constant)
2*L = 8 nm
mc^2 = 0.511×10^6 eV
Where m = mass electron
C= speed of light
(-2αL) = [-8nm/(197 eV.nm)] × (1.022× 10^6 eV*×3 eV)^0.5
(-2αL) = -71.1
Probability = e^(-2αL) = e^-71.1 = 1.3 ×10^-31
Therefore, the probability that a ondu tion ele tron in one wire arriving at the gap will pass through the gap into the other wire 1.3 ×10^-31.
Answer:
0.208 N
Explanation:
We are given that
Distance,d=0.41 m
The magnitude of the net electrostatic force experienced by any charge at point 4
Net force,
Where 
