Answer:
The energy difference between these 2p and 2s orbitals is
Explanation:
Wavelength of the photon emitted =
Energy of the photon will corresponds to the energy difference between 2p and 2s orbital = E
Energy of the photon is given by Planck's equation:
h = Planck's constant =
c = Speed of the light =
The energy difference between these 2p and 2s orbitals is
Answer : The cell emf for this cell is 0.118 V
Solution :
The half-cell reaction is:
In this case, the cathode and anode both are same. So, is equal to zero.
Now we have to calculate the cell emf.
Using Nernest equation :
where,
n = number of electrons in oxidation-reduction reaction = 1
= ?
= 0.0222 M
= 2.22 M
Now put all the given values in the above equation, we get:
Therefore, the cell emf for this cell is 0.118 V
