Question

If there are only two H-atoms, each is in 3rd excited state then:
A. Maximum number of different photons emitted is 4
B. Maximum number of different photons emitted is 3
C. Maximum number of different photons emitted is 1
D. Maximum number of different photons emitted is 2

Answer 1

Hint: The photon is a type of elementary particle. It is defined as the quantum of the electromagnetic field including electromagnetic radiation like light and radio waves and the force carrier for the electromagnetic force. Photons are generally massless, that's why they always move at the speed of light in vacuum.

Complete Step by step solution: Generally in the starting state known as ground state photons are present in this state. Now the question states that there are only two H-atoms, each is in 3rd excited state then number of photons would be:
As we know that number of photons for nth excited state can be calculated as:
$n=({{n}^{th}}+1)$
Here nth term = 3 given
$n=(3+1)=4$
This can be explained in the way that for the first H-atom the possible number of photons emitted in n = 4 to n = 3 then n = 3 to n = 2 after that n = 2 to n = 1 and for second H-atom the possible number of photons emitted from 4th excited state to 2nd i.e. from n = 4 to n = 2 and then from n = 2 to 1 which is repeated so we can say that the maximum number of photons emitted is 4.

Note: From ground state to move in higher state also known by excited state they need some source of energy that energy is calculated by the formula
$E=nh\nu $
Where E = energy, n = number of photons, h = Planck’s constant and $\nu $= frequency.