What is likely to be principal quantum number for the circular orbit of diameter 20.6 NM of the hydrogen atom if we ask him for Orbit to be the same as the represented by the principal quantum number is what is likely to be principal quantum number for a circular orbit of diameter 20.6 NM of the hydrogen atom if we ask him bohr Orbit to be the same as that represented by the principal quantum number is
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Secondary School Chemistry 13+7 pts
What is likely to be orbit number for a circular orbit of diameter 20 nm of the hydrogen atom?
Report by Chrisha7082 05.05.2019
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aqsaahmed19945
Aqsaahmed19945Virtuoso
Answer:
14 will likely to be the orbit number for a circular orbit.
Explanation:
Given Data :
Diameter = d = 20nm
radius = r = 10nm
r of nth shell= 0.539 X n X n
1nm = 0.1A= 0.1e -10
As,
0.539A = 0.0539 nm
Solution :
r = r of nth shell
10 = 0.0539 x n x n
n x n = 185.528
By taking its square root, we will get
n = 13.62
The orbit can't be 13.62. So, it's 14.( rounded off value)
● Answer -
Orbit number = 14
◆ Explaination -
# Given -
d = 20 nm
r = 10 nm = 100 A°
# Solution -
Bohr's radius of H-atom (in A°) is calculated by formula -
r = 0.529 n^2
100 = 0.529 n^2
n^2 = 100 / 0.529
n^2 = 189
Taking square root,
n = √189
n = 13.75
Rounding off, we can say that principal quantum number of given orbit is 14