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Answer:
(a) 0.3778 eV
(b) Ratio = 0.0278
Explanation:
The Bohr's formula for the calculation of the energy of the electron in nth orbit is:

(a) The energy of the electron in n= 6 excited state is:


Ionisation energy is the amount of this energy required to remove the electron. Thus, |E| = 0.3778 eV
(b) For first orbit energy is:




Ratio = 0.0278
Answer:
0.075 T
Explanation:
When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

where
I is the current in the wire
L is the length of the wire
B is the strength of the magnetic field
is the angle between the direction of I and B
In this problem we have:
L = 0.65 m is the length of the wire
I = 8.2 A is the current in the wire
F = 0.40 N is the force experienced by the wire
since the current is at right angle with the magnetic field
Solving the formula for B, we find the strength of the magnetic field:

Answer:
see below
Explanation:
a. 0.1886 x 12
=2.2632
This has 2 sig figures so the answer can only have 2 sig figures
2.3
b. 2.995 - 0.16685
=2.82815
The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place
2.828
c. 910 x 0.18945=172.3995
The least number of significant figures is 3 so the answer can only have 3 significant figures
172