Answer:
Explanation:
The Hi line of the Balmer series is emitted in the transition from n = 3 to n = 2 i.e. 
and 
The wavelength of Hi line of the Balmer series is given by :
So, the wavelength for this line is 550 nm. Hence, this is the required solution.
Answer:
Q1 = +2.50 x 10^-5C and Q2 = -2.50 x 10^-5C, r = 0.50m, F=?
Using Coulomb's law:
F = 1/(4πE) x Q1 x Q2/ r^2
Where
k= 1/(4πE) = 9 x 10^9Nm2/C2
Therefore,
F = 9x 10^9 x 2.50 x 10^-5 x2.50 x
10^-5/. ( 0.5)^2
F= 5.625/ 0.25
F= 22.5N approximately
F= 23N.
To find the direction of the force: since Q1 is positive and Q2 is negative, the force along Q1 and Q2 is force of attraction.
Hence To = 23N, attractive. C ans.
Thanks.
Normal force is mass x gravity, so mass x 9.81
Answer:
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