JEE MAIN - Chemistry (2004 - No. 57)
The wavelength of the radiation emitted when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097 $$\times$$ 107 m-1)
406 nm
192 nm
91 nm
9.1 $$\times$$ 10-8 nm
Explanation
We know Rydberg formula,
$${1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
[ for hydrogen atom Z = 1 ]
= $$1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$$
$$ \therefore $$ $$\lambda $$ = $${1 \over {1.097 \times {{10}^7}}}$$
= 9.11 $$ \times $$10-8 m
= 91.1 $$ \times $$10-9 m
= 91.1 nm
[ as 1 nm = 10-9 m ]
$${1 \over \lambda } = R \times {Z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
[ for hydrogen atom Z = 1 ]
= $$1.097 \times {10^7}\left( {{1 \over {{1^2}}} - {1 \over {{\infty ^2}}}} \right)$$
$$ \therefore $$ $$\lambda $$ = $${1 \over {1.097 \times {{10}^7}}}$$
= 9.11 $$ \times $$10-8 m
= 91.1 $$ \times $$10-9 m
= 91.1 nm
[ as 1 nm = 10-9 m ]
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