Work= force x displacement :)
Answer:
(a) 6.567 * 10^15 rev/s or hertz
(b) 8.21 * 10^14 rev/s or hertz
Explanation:
Fn= 4π^2k^2e^4m * z^2/(h^3*n^3)
Where Fn is frequency at all levels of n.
Z = 1 (nucleus)
e = 1.6 * 10^-19c
m = 9.1 * 10^-31 kg
h = 6.62 * 10-34
K = 9 * 10^9 Nm2/c2
(a) for groundstate n = 1
Fn = 4 * π^2 * (9*10^9)^2*(1.6*10^-19)^4* (9.1 * 10^-31) * 1 / (6.62 * 10^-31)^3 = 6.567 * 10^15 rev/s
(b) first excited state
n = 1
We multiple the groundstate answer by 1/n^3
6.567 * 10^15 rev/s/ 2^3
F2 = 8.2 * 10^ 14 rev/s
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To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.
The electric field in terms of the Force can be expressed as
Where,
F = Force
E= Electric Field
q = Charge
Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que
KE = W
KE = F*d
In the First Case,
In Second Case,
The total energy change would be subject to,
Therefore the Kinetic Energy change of the charged object is 27.976J
