Question

The surface of sodium metal is irradiated with radiation of wavelength \( x \) nm. The kinetic energy of ejected electrons is \( 2.8 \times 10^{-20} \) J. The work function of sodium is 2.3 eV. The value of \( x \) is _______ \( \times 10^2 \) nm. (Nearest integer)}

Answer 1

Correct Answer: 5

We are given the following data:
- The kinetic energy of ejected electrons: \( E_k = 2.8 \times 10^{-20} \) J,
- The work function of sodium: \( \phi = 2.3 \) eV,
- Planck's constant: \( h = 6.6 \times 10^{-34} \) J·s,
- The speed of light: \( c = 3.0 \times 10^8 \) m/s,
- The conversion factor: \( 1 \, \text{eV} = 1.6 \times 10^{-19} \) J.
The energy of a photon is given by the equation: \[ E_{\text{photon}} = \frac{hc}{x} \] where \( x \) is the wavelength of the radiation. According to the photoelectric equation: \[ E_{\text{photon}} = E_k + \phi \] Thus, the energy of the photon is the sum of the kinetic energy of the ejected electron and the work function: \[ \frac{hc}{x} = E_k + \phi \] Substitute the given values: \[ \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{x} = 2.8 \times 10^{-20} + (2.3 \times 1.6 \times 10^{-19}) \] Simplifying: \[ \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{x} = 2.8 \times 10^{-20} + 3.68 \times 10^{-19} \] \[ \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{x} = 4.0 \times 10^{-19} \] Now, solve for \( x \): \[ x = \frac{(6.6 \times 10^{-34})(3.0 \times 10^8)}{4.0 \times 10^{-19}} = 5.0 \times 10^{-7} \, \text{m} \] Convert this into nm: \[ x = 5.0 \times 10^{-7} \, \text{m} = 5.0 \times 10^2 \, \text{nm} \] Thus, the value of \( x \) is \( \boxed{5} \).