JEE MAIN - Chemistry (2020 - 4th September Evening Slot - No. 17)
The shortest wavelength of H atom in the
Lyman series is $$\lambda $$1. The longest wavelength in
the Balmar series of He+ is :
$${{5{\lambda _1}} \over 9}$$
$${{36{\lambda _1}} \over 5}$$
$${{27{\lambda _1}} \over 5}$$
$${{9{\lambda _1}} \over 5}$$
Explanation
For Shortest wavelength energy should be maximum.
For maximum energy transition must be form n = $$\infty $$ to n = 1.
$${1 \over {{\lambda _1}}} = {R_H}{\left( 1 \right)^2}\left[ {{1 \over 1} - 0} \right]$$
$$ \Rightarrow $$ $${1 \over {{\lambda _1}}} = {R_H}$$
For longest wavelength, $$\Delta $$E = minimum.
For Blamer series n = 3 to n = 2 will have $$\Delta $$E minimum for He+, Z = 2
So $${1 \over {{\lambda _2}}} = {R_H}{\left( 2 \right)^2}\left[ {{1 \over 4} - {1 \over 9}} \right]$$
$$ \Rightarrow $$ $${1 \over {{\lambda _2}}} = {R_H} \times {5 \over 9}$$
$$ \Rightarrow $$ $${\lambda _2} = {\lambda _1} \times {9 \over 5}$$
For maximum energy transition must be form n = $$\infty $$ to n = 1.
$${1 \over {{\lambda _1}}} = {R_H}{\left( 1 \right)^2}\left[ {{1 \over 1} - 0} \right]$$
$$ \Rightarrow $$ $${1 \over {{\lambda _1}}} = {R_H}$$
For longest wavelength, $$\Delta $$E = minimum.
For Blamer series n = 3 to n = 2 will have $$\Delta $$E minimum for He+, Z = 2
So $${1 \over {{\lambda _2}}} = {R_H}{\left( 2 \right)^2}\left[ {{1 \over 4} - {1 \over 9}} \right]$$
$$ \Rightarrow $$ $${1 \over {{\lambda _2}}} = {R_H} \times {5 \over 9}$$
$$ \Rightarrow $$ $${\lambda _2} = {\lambda _1} \times {9 \over 5}$$
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