The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectron

Question

The work function of a metal is 3 eV. The color of the visible light that is required to cause emission of photoelectrons is

Accepted Answer

Given: The work function of the metal is $\phi = 3 \, \text{eV}$ .

Find: The color of visible light required to emit photoelectrons.

For photoelectric emission, the incident photon must have energy at least equal to the work function. At threshold,

E = h\nu = \phi

Using the wavelength relation,

E = \frac{hc}{\lambda}

So,

\lambda = \frac{hc}{E}

Substituting $E = 3 \, \text{eV}$ , $h = 4.135667696 \times 10^{-15} \, \text{eV} \cdot \text{s}$ , and $c = 3 \times 10^8 \, \text{m/s}$ ,

\lambda = \frac{4.135667696 \times 10^{-15} \times 3 \times 10^8}{3}

\lambda \approx 414 \, \text{nm}

Using Einstein's photoelectric equation at threshold,

E = h\nu = \phi + KE_{\max}

At the threshold of emission,

KE_{\max} = 0

Hence,

h\nu_0 = \phi

Convert the work function into joules:

\phi = 3 \times 1.602 \times 10^{-19} \, \text{J} = 4.806 \times 10^{-19} \, \text{J}

Now,

\nu_0 = \frac{\phi}{h} = \frac{4.806 \times 10^{-19}}{6.626 \times 10^{-34}} \approx 7.25 \times 10^{14} \, \text{Hz}

The corresponding wavelength is

\lambda = \frac{c}{\nu_0} = \frac{3 \times 10^8}{7.25 \times 10^{14}} \approx 413 \, \text{nm}

This lies in the blue region of the visible spectrum. Therefore, the correct option is B.

Answer