When both sides of an equation give the same units, same numerical values, and same concept we refer to the equation as being correct. ... Removing constants from correct equations make them homogeneous but incorrect.
Answer:
a) 23.2 e V
b) energy of the original photon is 36.8 eV
Explanation:
given,
energy at ground level = -13.6 e V
energy at first exited state = - 3.4 e V
A photon of energy ionized from ground state and electron of energy K is released.
h ν₁ - 13.6 = K
K combine with photon in first exited state giving out photon of energy
= 26.6 e V
h c = 6.626 × 10⁻³⁴ × 3 × 10⁸ = 12400 e V A°
K + ( 3.4 ) = 26.6 e V
a) energy of free electron
K = 26.6 - 3.4 = 23.2 e V
b) energy of the original photon
h ν₁ - 13.6 = K
h ν₁ = 23.2 + 13.6
= 36.8 e V
energy of the original photon is 36.8 eV
In my opinion it does. The more water the pot holds, the longer you need to wait for it to freeze. Since there is more water, some parts may not be completely frozen. An experiment you can try is to get an ice cube container and a pot. fill both of them and put them in the freezer for the same amount of time. When you take it out, the ice cubes should be frozen leaving the pot with cold water.
epicycles were orbits within orbits used to explain discrepancies between expected and observed planetary movement, including the appearance of planets slowing down, speeding up, and moving backward.
Answer:
Mutual inductance, 
Explanation:
(a) A toroidal solenoid with mean radius r and cross-sectional area A is wound uniformly with N₁ turns. A second thyroidal solenoid with N₂ turns is wound uniformly on top of the first, so that the two solenoids have the same cross-sectional area and mean radius.
Mutual inductance is given by :

(b) It is given that,


Radius, r = 10.6 cm = 0.106 m
Area of toroid, 
Mutual inductance, 

or

So, the value of mutual inductance of the toroidal solenoid is
. Hence, this is the required solution.