Question

the maximum no. of emmision lines obtained when excited electron of a H- atoms in n=6 drops to the ground state is ______________​

Answer 1

$\Large\bold{\underline{\underline{Answer:-}}}$

$\Large\bold{\boxed{\boxed{15 }}}$$\rule{200}{4}$

$\bf\small\bold{\underline{\underline{Step-By-Step\:Explanation:-}}}$

Maximum number of spectral lines obatined when excited electron on 6th orbitals drop to ground state can be found as :

$\bf\bold{(n_2 - n_1) = (6 -1) = 5}$

Therefore, 5 transitions arising out of 6th level

Following transition can occur :

• 6→5, 6→4, 6→3, 6→2, 6→1 =5
• 5→4, 5→3, 5→2, 5→1 = 4
• 4→3, 4→2, 4→1 = 3
• 3→2, 3→1 = 2
• 2→1 = 1

Hence, Total Number of spectral lines = 5 + 4 + 3 + 2 + 1 = 15 lines.

Answer 2

HYDROGEN SPECTRUM :

Solution :

Total of emission lines will be ( 5 + 4 + 3 + 2 + 1 ) = 15

Number of spectral lines produced = $\boxed{\mathsf{\dfrac{n(n - 1)} {2}}}$

Number of spectral lines = $\mathsf{\dfrac{6 ( 6 - 1 )} {2}}$

Number of spectral lines = 15