The wavelengths involved in the spectrum of deuterium $\left({ }_1^2\mathrm{D}\right)$ are slightly different from that of hydrogen spectrum, because

Hint
Use the Rydberg formula to get the wavelength of the emitted photons and check if the wavelength of the emitted photon depends on the mass of the nuclei or any other factor.

Complete step by step answer
We know that the emission spectrum of atomic hydrogen has been divided into a number of spectral series with wavelengths given by the Rydberg formula.
According to the Rydberg formula, the energy differences between the levels in the Bohr model and the wavelengths of emitted photons is given by
As ${\displaystyle \frac{1}{\lambda }}=Z^{2}R\left({\displaystyle \frac{1}{n_1^2}}-{\displaystyle \frac{1}{n_2^2}}\right)$
Where Rydberg constant $R={\displaystyle \frac{1.097\times {10}^7}{1+{\displaystyle \frac{m}{M}}}}$here $m=$mass of the electron
$M=$ mass of the nucleus
$\lambda =$Wavelength of spectrum
$Z=$atomic number
$n_1$ $=$principal quantum number of the lower energy level.
 $n_2$$=$principal quantum number of the higher energy level.
Values are returned only when $n_1$ is less than $n_2$. For hydrogen spectral lines $Z=1$.
From the Rydberg formula, we can say that if Mass $M$ is different, $R$ is different, therefore the $\lambda$ value also changes.
Hence the wavelengths involved in the spectrum of deuterium $(D_1^2)$ are slightly different from that of the hydrogen spectrum because the masses of the two nuclei are different.
So the correct option is (C)