What is the maximum number of emission lines when the excited electron of a hydrogen atom in n = 6 drops to ground state?

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A 15
Maximum number of emission lines from $n^{t h}$ energy state to ground state is given by:

$= \frac{n (n - 1)}{2} = \frac{6 (6 - 1)}{2} = \frac{30}{2} = 15$ lines.

So, the correct option is $B$

Suggest Corrections

Similar questions

H - a t o m

n = 6

What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

What is the maximum number of emission lines when the excited electron of a H atom in n = 6 drops to the ground state?

Join BYJU'S Learning Program