Answer:
The energy of an electron in an isolated atom depends on b. n only.
Explanation:
The quantum number n, known as the principal quantum number represents the relative overall energy of each orbital.
The sets of orbitals with the same n value are often referred to as an electron shell, in an isolated atom all electrons in a subshell have exactly the same level of energy.
The principal quantum number comes from the solution of the Schrödinger wave equation, which describes energy in eigenstates
, and for the case of an hydrogen atom we have:
![E_n=-\cfrac{13.6}{n^2}\, eV](https://tex.z-dn.net/?f=E_n%3D-%5Ccfrac%7B13.6%7D%7Bn%5E2%7D%5C%2C%20eV)
Thus for each value of n we can describe the orbital and the energy corresponding to each electron on such orbital.
Answer:
t = 5.56 ms
Explanation:
Given:-
- The current carried in, Iin = 1.000002 C
- The current carried out, Iout = 1.00000 C
- The radius of sphere, r = 10 cm
Find:-
How long would it take for the sphere to increase in potential by 1000 V?
Solution:-
- The net charge held by the isolated conducting sphere after (t) seconds would be:
qnet = (Iin - Iout)*t
qnet = t*(1.000002 - 1.00000) = 0.000002*t
- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:
V = k*qnet / r
Where, k = 8.99*10^9 ..... Coulomb's constant
qnet = V*r / k
t = 1000*0.1 / (8.99*10^9 * 0.000002)
t = 5.56 ms
Mirrors reflect meaning shows what would be seen if everything was turned around, therefore, it shows text backwards because if it is turned around, it is in reverse order.
Answer:
![t=45.7s](https://tex.z-dn.net/?f=t%3D45.7s)
![\alpha =116revolutions](https://tex.z-dn.net/?f=%5Calpha%20%3D116revolutions)
Explanation:
Since we have given values of ω₀=32.o rad/s ,ω=0 and α=-0.700 rad/s² to find t we use below equation
![w=w_{o}+at\\ 0=(32.0rad/s)+(-0.700rad/s^{2} )t\\t=\frac{-32.0}{-0.700} \\t=45.7s](https://tex.z-dn.net/?f=w%3Dw_%7Bo%7D%2Bat%5C%5C%20%200%3D%2832.0rad%2Fs%29%2B%28-0.700rad%2Fs%5E%7B2%7D%20%29t%5C%5Ct%3D%5Cfrac%7B-32.0%7D%7B-0.700%7D%20%5C%5Ct%3D45.7s)
To find revolutions we use below equation
![w^{2}=w_{o}^{2}+2a\alpha](https://tex.z-dn.net/?f=w%5E%7B2%7D%3Dw_%7Bo%7D%5E%7B2%7D%2B2a%5Calpha)
Substitute the given values to find revolutions α
So
![0=(32.0rad/s)^{2}+2(-0.700rad/s^{2} )\alpha \\\alpha =\frac{(-32.0rad/s)^{2}}{2(-0.700rad/s^{2} )} \\\alpha =731rad](https://tex.z-dn.net/?f=0%3D%2832.0rad%2Fs%29%5E%7B2%7D%2B2%28-0.700rad%2Fs%5E%7B2%7D%20%29%5Calpha%20%20%5C%5C%5Calpha%20%3D%5Cfrac%7B%28-32.0rad%2Fs%29%5E%7B2%7D%7D%7B2%28-0.700rad%2Fs%5E%7B2%7D%20%29%7D%20%5C%5C%5Calpha%20%3D731rad)
To convert rad to rev:
![\alpha =(731rad)*(\frac{1rev}{2\pi rad} )\\\alpha =116revolutions](https://tex.z-dn.net/?f=%5Calpha%20%3D%28731rad%29%2A%28%5Cfrac%7B1rev%7D%7B2%5Cpi%20rad%7D%20%29%5C%5C%5Calpha%20%3D116revolutions)
Answer:
, ![2.18\cdot 10^6 m/s](https://tex.z-dn.net/?f=2.18%5Ccdot%2010%5E6%20m%2Fs)
Explanation:
The magnitude of the electromagnetic force between the electron and the proton in the nucleus is equal to the centripetal force:
![k\frac{(e)(e)}{r^2}=m_e \frac{v^2}{r}](https://tex.z-dn.net/?f=k%5Cfrac%7B%28e%29%28e%29%7D%7Br%5E2%7D%3Dm_e%20%5Cfrac%7Bv%5E2%7D%7Br%7D)
where
k is the Coulomb constant
e is the magnitude of the charge of the electron
e is the magnitude of the charge of the proton in the nucleus
r is the distance between the electron and the nucleus
v is the speed of the electron
is the mass of the electron
Solving for v, we find
![v=\sqrt{k\frac{e^2}{m_e r}}](https://tex.z-dn.net/?f=v%3D%5Csqrt%7Bk%5Cfrac%7Be%5E2%7D%7Bm_e%20r%7D%7D)
Inside an atom of hydrogen, the distance between the electron and the nucleus is approximately
![r=5.3\cdot 10^{-11}m](https://tex.z-dn.net/?f=r%3D5.3%5Ccdot%2010%5E%7B-11%7Dm)
while the electron mass is
![m_e = 9.11\cdot 10^{-31}kg](https://tex.z-dn.net/?f=m_e%20%3D%209.11%5Ccdot%2010%5E%7B-31%7Dkg)
and the charge is
![e=1.6\cdot 10^{-19} C](https://tex.z-dn.net/?f=e%3D1.6%5Ccdot%2010%5E%7B-19%7D%20C)
Substituting into the formula, we find
![v=\sqrt{(9\cdot 10^9 m/s) \frac{(1.6\cdot 10^{-19} C)^2}{(9.11\cdot 10^{-31} kg)(5.3\cdot 10^{-11} m)}}=2.18\cdot 10^6 m/s](https://tex.z-dn.net/?f=v%3D%5Csqrt%7B%289%5Ccdot%2010%5E9%20m%2Fs%29%20%5Cfrac%7B%281.6%5Ccdot%2010%5E%7B-19%7D%20C%29%5E2%7D%7B%289.11%5Ccdot%2010%5E%7B-31%7D%20kg%29%285.3%5Ccdot%2010%5E%7B-11%7D%20m%29%7D%7D%3D2.18%5Ccdot%2010%5E6%20m%2Fs)