QJEEM18Ph1a1a: JEE Main 2018 Physics - Class 12 JEEM18Ph1a1a Solution: If the series limit frequency of the Lyman series is , then the …

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,

${1 \over \lambda }$   =  RZ² $\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$

Series limit means transition happens

from n = $\propto$ to n = 1, for Lyman Series.

In series limit for Lyman series,

${1 \over {{\lambda _L}}}$ = RZ² $\left( {{1 \over {{1^2}}} - {1 \over \propto }} \right)$

$\Rightarrow$ $\,\,\,$ ${1 \over {{\lambda _L}}}$ = RZ²

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 $\times$ RZ²

In Pfund series,

n₂ = $\propto$  and  n₁ = 5

$\therefore\,\,\,$ ${1 \over {{\lambda _P}}}$  = RZ² $\left( {{1 \over {{5^2}}} - {1 \over {{ \propto ^2}}}} \right)$

$\Rightarrow$ $\,\,\,$ ${1 \over {{\lambda _P}}}$ = ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ ${\gamma _P}$   =   ${c \over {{\lambda _P}}}$ = c $\times$ ${{R{Z^2}} \over {25}}$

$\therefore\,\,\,$ $\gamma$ _P = ${{cRZ{}^2} \over {25}}$ = ${{{\gamma _L}} \over {25}}$    [as   $\gamma$ _L = cRZ²]