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

$${1 \over \lambda } = {R_H}{Z^2}\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]$$

$$\therefore$$ $${1 \over {91.2}} = {R_H}\left[ {{1 \over {{1^2}}} - {1 \over {{n^2}}}} \right]$$

Since R_H = 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