Question

What would happen to the energy levels of a hydrogen atom if the Coulomb force doubled in strength? What would happen to the sizes of atoms?

Short answer

Answer: When the Coulomb force doubles, the energy levels of a hydrogen atom will become four times more negative (lower), and the size of the atom will be reduced to half of its original size.

Step-by-step solution

Recall the energy level formula for a hydrogen atom

The energy level formula for a hydrogen atom according to Bohr's model is: E_n = - \frac{Z^2 e^4 m_e}{2 (4 \pi \epsilon_0)^2 \hbar^2 n^2} where E_n is the energy of the nth orbit (energy level), Z is the atomic number, e is the electron charge, m_e is the mass of an electron, \epsilon_0 is the vacuum permittivity, \hbar is the reduced Planck constant, and n is the principal quantum number.

Double the strength of the Coulomb force

Doubling the Coulomb force means that the value of e^2 will be doubled, so we represent the Coulomb force constant as 2e^2. The new energy level formula when the Coulomb force is doubled will thus be: E'_n = - \frac{Z^2 (2e^2)^2 m_e}{2 (4 \pi \epsilon_0)^2 \hbar^2 n^2}

Simplify the new energy level formula and compare it with the original formula

Simplifying the new energy level formula, we get: E'_n = - \frac{4 Z^2 e^4 m_e}{2 (4 \pi \epsilon_0)^2 \hbar^2 n^2} Comparing this new formula with the original formula, the ratio of the new energy to the old energy is: \frac{E'_n}{E_n} = \frac{- \frac{4 Z^2 e^4 m_e}{2 (4 \pi \epsilon_0)^2 \hbar^2 n^2}}{- \frac{Z^2 e^4 m_e}{2 (4 \pi \epsilon_0)^2 \hbar^2 n^2}} = 4

Calculate the change in energy levels

Since the ratio of the new energy to the old energy is 4, it means that when the Coulomb force doubles, the energy levels of a hydrogen atom will become four times more negative (4 times lower).

Recall the formula for the size of a hydrogen atom

The size or radius (r) of a hydrogen atom according to Bohr's model is proportional to the square of the principal quantum number (n^2). The formula for the size of a hydrogen atom is: r_n = \frac{n^2 \hbar^2}{Z m_e e^2} (4 \pi \epsilon_0)

Substitute the doubled Coulomb force into the size formula and compare the sizes

When the Coulomb force is doubled, we have 2e^2 as before. Substitute this value into the size formula: r'_n = \frac{n^2 \hbar^2}{Z m_e (2e^2)} (4 \pi \epsilon_0) Comparing the new size formula with the original formula, we find the ratio of new size to old size: \frac{r'_n}{r_n}= \frac{\frac{n^2 \hbar^2}{Z m_e (2e^2)} (4 \pi \epsilon_0)}{\frac{n^2 \hbar^2}{Z m_e e^2} (4 \pi \epsilon_0)} = \frac{1}{2}

Calculate the change in the size of atoms

Since the ratio of the new size to the old size is \frac{1}{2}, it means that when the Coulomb force doubles, the size of the hydrogen atom will be halved. In conclusion, when the Coulomb force doubles, the energy levels of a hydrogen atom will become four times more negative (lower), and the size of the atom will be reduced to half of its original size.