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

The right solution is:

(a) 2.87 eV

(b) 1.4375 eV

Step-by-step explanation:

Given:

Wavelength,

= 433 nm

Potential difference,

= 1.43 V

Now,

(a)

The energy of photon will be:

E =
`(6.626* 10^(-34)* 3* 10^8)/(433* 10^(-9))`

=
`4.59* 10^(-19) \ J`

or,

=
`(4.59* 10^(-19))/(1.6* 10^(-19))`

=
`2.87 \ eV`

(b)

As we know,

⇒
`Vq=(hc)/(\lambda)-\Phi_0`

By substituting the values, we get

⇒
`1.43* 1.6* 10^(19)=(6.626* 10^(-34)* 3* 10^8)/(433* 10^(-9))-\Phi_0`

⇒
`\Phi_0=2.3* 10^(-19) \ J`

or,

⇒
`=(2.3* 10^(-19))/(1.6* 10^(-19))`

⇒
`=1.4375 \ eV`