D

In the Bohr model of the hydrogen atom, the energy levels and radius for the electron can be derived from the formulas:
Energy: The energy of the electron in the nth orbit is given by:
$$ E_n = - \frac{m e^4}{8 \epsilon_0^2 h^2 n^2} $$
Where:

• m is the mass of the electron,
• e is the charge of the electron,
• \( \epsilon_0 \) is the permittivity of free space,
• h is Planck's constant,
• n is the principal quantum number.

In this case, if the mass of the electron is doubled (m becomes 2m), then the new energy becomes:
$$ E_n = - \frac{(2m) e^4}{8 \epsilon_0^2 h^2 n^2} = -2 \left( \frac{m e^4}{8 \epsilon_0^2 h^2 n^2} \right) = 2E_0 $$
Where \( E_0 = -27.2 ext{ eV} \). Thus, the new energy in the first orbit will be \( E_1 = 2(-27.2) = -54.4 ext{ eV} \).
Radius: The radius of the nth orbit can be expressed as:
$$ r_n = \frac{n^2 h^2 \epsilon_0}{\pi m e^2} $$
Again, doubling the mass (m becomes 2m) results in:
$$ r_n = \frac{n^2 h^2 \epsilon_0}{\pi (2m) e^2} = \frac{1}{2} \left( \frac{n^2 h^2 \epsilon_0}{\pi m e^2} \right) = \frac{r_0}{2} $$
Hence, the new radius in the first orbit becomes \( r_1 = \frac{a_0}{2} \).
Putting together the results:

• New energy is \( -54.4 ext{ eV} \),
• New radius is \( \frac{a_0}{2}. \)

Thus, the final answers are option D: \( 2E_0 \) and \( r_0 = \frac{a_0}{2} \).