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:

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 =

  =

or,

  =

  =

(b)

As we know,

⇒

By substituting the values, we get

⇒

⇒                      

or,

⇒                            

⇒                            

6/12=1/2

step-by-step explanation:

answer: good job it's all correct !

explanation: