First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5
The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm
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Answer:
on the surface of the cathode
Answer: n=15.56moles
Explanation:
PV = nRT
where
P is pressure in atmospheres
V is volume in Liters
n is the number of moles of the gas
R is the ideal gas constant = given as (0.0821L -atm/k-mol
PV = nRT
n= PV/RT
n= (1.5 X 230)/ (0.0821 X 270)
n= 15.56 moles
