Question

When you irradiate a metal with light of wavelength 433 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.43 V is needed to reduce the current to zero. What is the energy of a photon of this light in electron volts? energy of a photon: Find the work function of the irradiated metal in electron volts. work function:

Answer 1

Answer:

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E = $\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}$

  = $4.59\times 10^{-19} \ J$

or,

  = $\frac{4.59\times 10^{-19}}{1.6\times 10^{-19}}$

  = $2.87 \ eV$

(b)

As we know,

⇒ $Vq=\frac{hc}{\lambda}-\Phi_0$

By substituting the values, we get

⇒ $1.43\times 1.6\times 10^{19}=\frac{6.626\times 10^{-34}\times 3\times 10^8}{433\times 10^{-9}}-\Phi_0$

⇒                       $\Phi_0=2.3\times 10^{-19} \ J$

or,

⇒                            $=\frac{2.3\times 10^{-19}}{1.6\times 10^{-19}}$

⇒                            $=1.4375 \ eV$