Question

An electron makes transition from $n=4$ to $n=1$ state in a hydrogen atom. The maximum possible number of photons emitted will be :

• A. $1$
• B. $2$
• C. $3$
• D. $6$

Answer 1

The correct option is D $6$

From the nth state, the electron may go to $(n-1{)}^{th}$ state, ...., 2nd state or 1st state. So there are (n 1) possible transitions starting from the $n^{th}$ state. The atoms reaching $(n-1{)}^{th}$ state may make (n 2) different transitions. The atoms reaching $(n-2{)}^{th}$ state may make (n 3) different transitions. Similarly for other lower states. During each transition a photon with energy $h\nu$ and wavelength $\lambda$ is emitted out. Hence, the total number of possible transitions is equal to the number of photons emitted. The total number of possible transitions is

$(n-1),(n-2),(n-3),................,3,2,1=\frac{n(n-1)}{2}$

Therefore, for transition of an electron from higher energy state n = 4 to lower energy state n = 1 the number of photons emitted are

$\frac{n(n-1)}{2}=\frac{4(4-1)}{2}=\frac{12}{2}=6$