QJEEM17Chad62: JEE Main 2017 Chemistry - Class 11 JEEM17Chad62 Solution: If the shortest wavelength in Lyman series of hydrogen atom is…

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 ( ${1 \over {n_1^2}}$ $-$ ${1 \over {n_2^2}}$ )

The shortest wavelength of hydrogen atom in Lyman series is from n₁ = 1 to n₂ = $\propto$

$\therefore\,\,\,$ ${1 \over A}$ = 1² R ( ${1 \over {{1^2}}}$ $-$ ${1 \over {{ \propto ^2}}}$ )    [for hydrogen, z = 1]

$\Rightarrow$ $\,\,\,$ ${1 \over A}$ = R

The longest wavelength in pascal series of He⁺ is from n₁ = 3 to n₂ = 4

For He⁺, z = 2.

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

$\Rightarrow$ $\,\,\,$ ${1 \over \lambda }$ = ${1 \over A}$ (2)² ( ${1 \over {{3^2}}}$ $-$ ${1 \over {{4^2}}}$ )

$\Rightarrow$ $\,\,\,$ ${1 \over \lambda }$ = ${4 \over A}$ $\left( {{7 \over {16 \times 9}}} \right)$ = ${7 \over {36A}}$

$\Rightarrow$ $\,\,\,$ $\lambda$ = ${{36A} \over 7}$