JEE MAIN - Chemistry (2019 - 11th January Morning Slot - No. 7)
Heat treatment of muscular pain involves radiation of wavelength of about 900 nm. Which spectral line of H atom is suitable for this purpose?
[RH = 1 $$ \times $$ 105 cm–1, h = 6.6 $$ \times $$ 10–34 Js, c = 3 $$ \times $$ 108 ms–1]
[RH = 1 $$ \times $$ 105 cm–1, h = 6.6 $$ \times $$ 10–34 Js, c = 3 $$ \times $$ 108 ms–1]
Balmer, $$\infty $$ $$ \to $$ 2
Paschen, 5 $$ \to $$ 3
Paschen, $$\infty $$ $$ \to $$ 3
Lyman, $$\infty $$ $$ \to $$ 1
Explanation
Given, RH = 1 $$ \times $$ 105 cm–1
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10-5 cm
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10-7 cm $$ \times $$ 100
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 100 nm
We know,
$${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$$
$$ \Rightarrow $$ $$\lambda $$ = $${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
[For H atom Z = 1]
$$ \Rightarrow $$ $$\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
$$ \Rightarrow $$ $$\lambda $$ = $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
Given, $$\lambda $$ = 900 nm
$$ \therefore $$ $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$ = 900
$$ \Rightarrow $$ $${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$
By checking each options you can see
when nL = 3 and nH = $$\infty $$ then
$${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$
$$ \therefore $$ Option C is correct.
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10-5 cm
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 10-7 cm $$ \times $$ 100
$$ \Rightarrow $$ $${1 \over {{R_H}}}$$ = 100 nm
We know,
$${1 \over \lambda } = \nu = {R_H} \times {Z^2}\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)$$
$$ \Rightarrow $$ $$\lambda $$ = $${1 \over {{R_H} \times {{\left( 1 \right)}^2}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
[For H atom Z = 1]
$$ \Rightarrow $$ $$\lambda = {1 \over {{R_H}}} \times {1 \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
$$ \Rightarrow $$ $$\lambda $$ = $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$
Given, $$\lambda $$ = 900 nm
$$ \therefore $$ $${{100} \over {\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right)}}$$ = 900
$$ \Rightarrow $$ $${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$
By checking each options you can see
when nL = 3 and nH = $$\infty $$ then
$${\left( {{1 \over {n_L^2}} - {1 \over {n_H^2}}} \right) = {1 \over 9}}$$
$$ \therefore $$ Option C is correct.
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