Question

There are only three hydrogen atoms in a discharge tube. The analysis of spectrum shows that in all the hydrogen atoms, electrons are de-exciting from the fourth orbit. What should be the maximum number of spectral lines?

• A. $6$
• B. $1$
• C. $4$
• D. $5$

Answer 1

The correct option is A $6$

Since, from the $n^{th}$ state, the electron may go to $(n-1{)}^{th}$ state,$(n-2{)}^{th}$ 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 are

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

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

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

Here, the electrons of each hydrogen atom are in same state hence in transition will emit photons of equal wavelengths for each transition hence, the maximum spectral lines emitted are $6$.