Question

Minimum energy required to remove an atom from sodium is \( 3.313 \times 10^{-19} \text{g.} \) Maximum wavelength of radiation that will get photoelectron

Hint:

To save calculation time in exams, memorize the value of \( hc \approx 1240 \text{ eV}\cdot\text{nm} \) or \( hc \approx 19.89 \times 10^{-26} \text{ J}\cdot\text{m} \).
Also, observe that \( 3.313 \) is exactly half of \( 6.626 \), making the division trivial without a calculator!

Answer 1

Step 1: Understanding the Concept:
The minimum energy required to remove an electron from a metal surface is called its work function (\( \Phi \)).
The maximum wavelength (threshold wavelength, \( \lambda_0 \)) capable of ejecting a photoelectron corresponds exactly to this minimum energy.
Step 2: Key Formula or Approach:
The energy of a photon relates to its wavelength by the Planck-Einstein relation:
\[ E = \frac{hc}{\lambda} \] Where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \text{ J s} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \text{ m/s} \)).
Note: The unit in the question is mistakenly typed as "g.", but it implies Joules (J).
Step 3: Detailed Explanation:
Given the work function (minimum energy) \( E = 3.313 \times 10^{-19} \text{ J} \).
We rearrange the formula to solve for the maximum wavelength \( \lambda \):
\[ \lambda = \frac{hc}{E} \] Substitute the known constants into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34} \text{ J s} \times 3 \times 10^8 \text{ m/s}}{3.313 \times 10^{-19} \text{ J}} \] First, multiply the numerator:
\[ hc \approx 19.878 \times 10^{-26} \text{ J m} \] Now, divide by the energy:
\[ \lambda = \frac{19.878 \times 10^{-26}}{3.313 \times 10^{-19}} \] Notice that \( \frac{19.878}{3.313} \approx 6.00 \).
\[ \lambda = 6.00 \times 10^{-7} \text{ m} \] To convert meters to nanometers (\( 1 \text{ nm} = 10^{-9} \text{ m} \)):
\[ \lambda = 600 \times 10^{-9} \text{ m} = 600 \text{ nm} \] Step 4: Final Answer:
The maximum wavelength of radiation is 600 nm.