JEE MAIN - Physics (2021 - 26th February Morning Shift - No. 18)
If $$\lambda$$1 and $$\lambda$$2 are the wavelengths of the third member of Lyman and first member of the Paschen series respectively, then the value of $$\lambda$$1 : $$\lambda$$2 is :
7 : 135
7 : 108
1 : 9
1 : 3
Explanation
For Lyman series
n1 = 1, n2 = 4
$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right]$$
For paschen series
n1 = 3, n2 = 4
$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]$$
$$ \therefore $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left[ {{1 \over 9} - {1 \over {16}}} \right]} \over {\left[ {1 - {1 \over {16}}} \right]}} = {7 \over {9 \times 15}}$$
$$ \Rightarrow $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {7 \over {135}}$$
n1 = 1, n2 = 4
$${1 \over {{\lambda _1}}} = R\left[ {{1 \over {{1^2}}} - {1 \over {{4^2}}}} \right]$$
For paschen series
n1 = 3, n2 = 4
$${1 \over {{\lambda _2}}} = R\left[ {{1 \over {{3^2}}} - {1 \over {{4^2}}}} \right]$$
$$ \therefore $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {{\left[ {{1 \over 9} - {1 \over {16}}} \right]} \over {\left[ {1 - {1 \over {16}}} \right]}} = {7 \over {9 \times 15}}$$
$$ \Rightarrow $$ $${{{\lambda _1}} \over {{\lambda _2}}} = {7 \over {135}}$$
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