Noble gases
also known as Inert gases
Answer : The energy of the photon emitted is, -12.1 eV
Explanation :
First we have to calculate the
orbit of hydrogen atom.
Formula used :

where,
= energy of
orbit
n = number of orbit
Z = atomic number of hydrogen atom = 1
Energy of n = 1 in an hydrogen atom:

Energy of n = 2 in an hydrogen atom:

Energy change transition from n = 1 to n = 3 occurs.
Let energy change be E.

The negative sign indicates that energy of the photon emitted.
Thus, the energy of the photon emitted is, -12.1 eV
Answer:
the atomic mass number for boron is 5
Explanation:
hope it helps , pls mark me as brainliest
Solving part-1 only
#1
KMnO_4
- Transition metal is Manganese (Mn)
#2
Actually it's the oxidation number of Mn
Let's find how?




- x is the oxidation number
#3
- Purple as per the color of potassium permanganate
#4
