Question

An electromagnetic radiation of wavelength 331.5 nm is made to strike the surface of a metal. Electrons are emitted with a kinetic energy of \(1.2 \times 10^5\) J mol\(^{-1}\). The work function (in eV) of the metal is (h=\(6.63 \times 10^{-34}\) Js, \(N_A = 6 \times 10^{23}\) mol\(^{-1}\))

• A. 1.5
• B. 3.0
• C. 3.5
• D. 2.5

Hint:

The photoelectric effect equation applies to a single photon interacting with a single electron. If any energy is given per mole, you must divide by Avogadro's number to get the per-particle energy before using the equation. A useful shortcut for photon energy is \(E(\text{eV}) = \frac{1240}{\lambda(\text{nm})}\). Here, \(E = 1240/331.5 \approx 3.74\) eV.

Answer 1

The Correct Option is D

Step 1: Use the photoelectric effect equation.
Einstein's photoelectric equation is \( E_{photon} = \phi + K.E._{max} \), where \(\phi\) is the work function and \(K.E._{max}\) is the maximum kinetic energy of an emitted electron. All energies must be for a single particle.
Step 2: Calculate the energy of a single photon (\(E_{photon}\)).
\( E_{photon} = \frac{hc}{\lambda} \), where \(h\) is Planck's constant, \(c\) is the speed of light (\(3 \times 10^8\) m/s), and \(\lambda\) is the wavelength.
Given \( \lambda = 331.5 \text{ nm} = 331.5 \times 10^{-9} \) m.
\( E_{photon} = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^8)}{331.5 \times 10^{-9}} = \frac{19.89 \times 10^{-26}}{331.5 \times 10^{-9}} \approx 0.06 \times 10^{-17} = 6 \times 10^{-19} \) J.
Step 3: Convert the given kinetic energy from J/mol to J/electron.
The given K.E. is for one mole of electrons: \( K.E._{mol} = 1.2 \times 10^5 \) J/mol.
To find the K.E. for a single electron, we divide by Avogadro's number, \(N_A\).
\( K.E._{electron} = \frac{K.E._{mol}}{N_A} = \frac{1.2 \times 10^5}{6 \times 10^{23}} = 0.2 \times 10^{-18} = 2 \times 10^{-19} \) J.
Step 4: Calculate the work function (\(\phi\)) in Joules.
\( \phi = E_{photon} - K.E._{electron} = (6 \times 10^{-19}) - (2 \times 10^{-19}) = 4 \times 10^{-19} \) J.
Step 5: Convert the work function from Joules to electron-volts (eV).
The conversion factor is \( 1 \text{ eV} = 1.6 \times 10^{-19} \) J.
\( \phi (\text{in eV}) = \frac{\phi (\text{in J})}{1.6 \times 10^{-19}} = \frac{4 \times 10^{-19}}{1.6 \times 10^{-19}} = \frac{4}{1.6} = \frac{40}{16} = \frac{5}{2} = 2.5 \) eV.