JEE MAIN - Physics (2003 - No. 2)
The wavelengths involved in the spectrum of deuterium $$\left( {{}_1^2\,D} \right)$$ are slightly different from that of hydrogen spectrum, because
the size of the two nuclei are different
the nuclear forces are different in the two cases
the masses of the two nuclei are different
the attraction between the electron and the nucleus is different in the two cases
Paskaidrojums
The wavelength of spectrum is given by
$${1 \over \lambda } = R{z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
where $$R = {{1.097 \times {{10}^7}} \over {1 + {m \over M}}}$$
where $$m=$$ mass of electron
$$M=$$ mass of nucleus.
For different $$M,R$$ is different and therefore $$\lambda $$ is different
$${1 \over \lambda } = R{z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$
where $$R = {{1.097 \times {{10}^7}} \over {1 + {m \over M}}}$$
where $$m=$$ mass of electron
$$M=$$ mass of nucleus.
For different $$M,R$$ is different and therefore $$\lambda $$ is different