Question

The work function of a certain metal is \(3.31 \times 10^{-19}\,J\). Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength \(5000\,\AA\) is (given \(h = 6.62 \times 10^{-34}\,Js\), \(c = 3 \times 10^8\,m\,s^{-1}\), \(e = 1.6 \times 10^{-19}\,C\))

• A. \(2.48\,eV\)
• B. \(0.41\,eV\)
• C. \(2.07\,eV\)
• D. \(0.82\,eV\)

Hint:

Always convert wavelength from \(\AA\) to meters. Use \(K_{max} = \frac{hc}{\lambda} - \phi\) and then convert joule to eV by dividing by \(1.6\times 10^{-19}\).

Answer 1

The Correct Option is B

Step 1: Use Einstein’s photoelectric equation.
\[ K_{max} = \frac{hc}{\lambda} - \phi \]
Step 2: Convert wavelength.
\[ \lambda = 5000\AA = 5000 \times 10^{-10}m = 5\times 10^{-7}m \]
Step 3: Calculate photon energy.
\[ E = \frac{hc}{\lambda} = \frac{(6.62\times 10^{-34})(3\times 10^8)}{5\times 10^{-7}} \]
\[ E = \frac{19.86\times 10^{-26}}{5\times 10^{-7}} = 3.972\times 10^{-19}J \]
Step 4: Subtract work function.
\[ K_{max} = 3.972\times 10^{-19} - 3.31\times 10^{-19} = 0.662\times 10^{-19}J \]
Step 5: Convert into eV.
\[ K_{max} = \frac{0.662\times 10^{-19}}{1.6\times 10^{-19}} = 0.414\,eV \approx 0.41\,eV \]
Final Answer:
\[ \boxed{0.41\,eV} \]