Question 292: Atoms & Nuclei

Question

If the series limit frequency of the Lyman series is $${
u _L}$$, then the series limit frequency of the Pfund series is:

Accepted Answer

Note
:
(1)   In Lyman Series,  transition happens in n = 1 state
from n = 2, 3, . . . . . $$ \propto $$
(2)   In Balmer Series,  transition happens in n = 2 state
from n = 3, 4, . . . . . $$ \propto $$
(3)   In Paschen Series,  transition happens in n = 3 state
from  n = 4, 5, . . . . . $$ \propto $$
(4)   In Bracktt Series, transition happens in n = 4 state
from n = 5, 6 . . . . . . $$ \propto $$
(5)   In Pfund Series, transition happens in n = 5 state
from n = 6, 7, . . . . $$ \propto $$
We know,
$${1 \over \lambda }$$  =  RZ
2
$$\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} ight)$$
Series limit means transition happens
from n = $$ \propto $$ to n = 1,  for Lyman Series.
In series limit for Lyman series,
$${1 \over {{\lambda _L}}}$$ = RZ
2
$$\left( {{1 \over {{1^2}}} - {1 \over  \propto }} ight)$$
$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {{\lambda _L}}}$$ = RZ
2
We know,
E = $${{hc} \over \lambda }$$ = h$$\gamma $$
$$ \Rightarrow $$$$\,\,\,$$ $$\gamma $$ = $${c \over \lambda }$$
So, frequency in Lyman Series,
$$\gamma $$
L
=  $${c \over {{\lambda _L}}}$$  = c $$ 	imes $$ RZ
2
In Pfund series,
n
2
= $$ \propto $$  and  n
1
= 5
$$	herefore\,\,\,$$ $${1 \over {{\lambda _P}}}$$ = RZ
2
$$\left( {{1 \over {{5^2}}} - {1 \over {{ \propto ^2}}}} ight)$$
$$ \Rightarrow $$$$\,\,\,$$ $${1 \over {{\lambda _P}}}$$ = $${{R{Z^2}} \over {25}}$$
$$	herefore\,\,\,$$ $${\gamma _P}$$    =   $${c \over {{\lambda _P}}}$$ = c $$ 	imes $$ $${{R{Z^2}} \over {25}}$$
$$	herefore\,\,\,$$ $$\gamma $$
P
=  $${{cRZ{}^2} \over {25}}$$ = $${{{\gamma _L}} \over {25}}$$    [as   $$\gamma $$
L
= cRZ
2
]

Answer

$${
u _L}/25$$

Answer

$$25{
u _L}$$

Answer

$$16{
u _L}$$

Answer

$${
u _L}/16$$

Atoms & Nuclei

If the series limit frequency of the Lyman series is ${\nu _L}$, then the series limit frequency of the Pfund series is:

Choose the correct answer: