JEE Advance - Chemistry (2003 - No. 4)
Wavelength of high energy transition of H-atoms is 91.2 nm. Calculate the corresponding wavelength of He atoms.
91.2 nm
45.6 nm
182.4 nm
22.8 nm
11.4 nm
Explanation
$${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$
$$\therefore$$ $${1 \over {91.2}} = {R_H}\left[ {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right]$$
Since RH = constant
$${{{\lambda _{He}}} \over {{\lambda _H}}} = {{Z_H^2} \over {Z_{He}^2}} = {1 \over 4}$$
or, $${\lambda _{He}} = {{{\lambda _H}} \over 4} = {{91.2} \over 4} = 22.8$$ nm
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