The photoelectric work function for a metal surface is 4.125 eV. The cut-off wavelength for this surface is

Question

A photon and an electron have the same energy \( E \). If \( \lambda_p \) is the wavelength of the photon and \( \lambda_e \) is the de Broglie wavelength of the electron, then the ratio \( \frac{\lambda_p}{\lambda_e} \) is:

Answer 1

The problem is related to the photoelectric effect, which describes the emission of electrons from a metal surface when it is exposed to light. The energy required to remove an electron from the surface of a metal is known as the work function (\phi). This work function is given as 4.125 eV for this problem.

To find the cut-off wavelength (\lambda_c), we use the photoelectric equation relating the work function and cut-off wavelength:

\phi = \frac{hc}{\lambda_c}

where:

h is Planck's constant (6.626 \times 10^{-34} \, \text{J s})
c is the speed of light (3 \times 10^8 \, \text{m/s})
\lambda_c is the cut-off wavelength
\phi is the work function given in electron volts (1 eV = 1.6 \times 10^{-19} \, \text{J})

First, convert the work function from electron volts to joules:

\phi = 4.125 \, \text{eV} = 4.125 \times 1.6 \times 10^{-19} \, \text{J}

Now, solve for the cut-off wavelength:

\lambda_c = \frac{hc}{\phi}

Substitute the values:

\lambda_c = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{4.125 \times 1.6 \times 10^{-19}}

Simplifying this expression:

\lambda_c = \frac{19.878 \times 10^{-26}}{6.6 \times 10^{-19}} = 3.012 \times 10^{-7} \, \text{m}

Convert meters to angstroms (1 m = 10^{10} \, \mathring{A}):

\lambda_c = 3.012 \times 10^{-7} \times 10^{10} = 3012 \, \mathring{A}

Rounding to the nearest option provided gives 3000 \, \mathring{A}, which matches the correct answer.

Therefore, the cut-off wavelength for this surface is 3000 \mathring{A}.

Answer 2