Questions: An X-ray photon of wavelength 150 pm ejects an electron from the inner part of an atom. The speed of the electron was measured and found to be 2.14 × 10^7 m / s. How tightly was it bound in the atom, in J?

Solution Steps

First, we need to calculate the energy of the X-ray photon using its wavelength. The energy \( E \) of a photon is given by the equation:

\[ E = \frac{hc}{\lambda} \]

where:

• \( h \) is Planck's constant (\( 6.6261 \times 10^{-34} \) J·s),
• \( c \) is the speed of light (\( 3.00 \times 10^{8} \) m/s),
• \( \lambda \) is the wavelength of the photon (150 pm = \( 150 \times 10^{-12} \) m).

Substituting the values:

\[ E = \frac{(6.6261 \times 10^{-34} \, \text{J·s}) (3.00 \times 10^{8} \, \text{m/s})}{150 \times 10^{-12} \, \text{m}} \]

\[ E = \frac{1.9878 \times 10^{-25} \, \text{J·m}}{150 \times 10^{-12} \, \text{m}} \]

\[ E = 1.3252 \times 10^{-15} \, \text{J} \]

The kinetic energy \( K \) of the ejected electron can be calculated using its speed \( v \):

\[ K = \frac{1}{2}mv^2 \]

where:

• \( m \) is the mass of the electron (\( 9.1094 \times 10^{-31} \) kg),
• \( v \) is the speed of the electron (\( 2.14 \times 10^{7} \) m/s).

Substituting the values:

\[ K = \frac{1}{2} (9.1094 \times 10^{-31} \, \text{kg}) (2.14 \times 10^{7} \, \text{m/s})^2 \]

\[ K = \frac{1}{2} (9.1094 \times 10^{-31}) (4.5796 \times 10^{14}) \]

\[ K = \frac{1}{2} (4.1726 \times 10^{-16}) \]

\[ K = 2.0863 \times 10^{-16} \, \text{J} \]

The binding energy \( E_b \) of the electron in the atom is the difference between the energy of the X-ray photon and the kinetic energy of the ejected electron:

\[ E_b = E - K \]

Substituting the values:

\[ E_b = 1.3252 \times 10^{-15} \, \text{J} - 2.0863 \times 10^{-16} \, \text{J} \]

\[ E_b = 1.1166 \times 10^{-15} \, \text{J} \]

Final Answer