Question

The work function of a certain metal is 6.6 eV. When a photon whose frequency is 2.80 x 1015 Hz is absorbed by an electron below the surface, the electron escapes the surface with 3.5 eV of kinetic energy. How much energy (in eV) did the electron use to reach the surface

Answer 1

The energy used to reach the surface is 1.53 eV.

Step-by-step explanation:

Given that, the work function of a certain metal is 6.6 eV.

Frequency of photon,
`f=2.8* 10^(15)\ Hz`

Kinetic energy of the electron,
`E=3.5\ eV`

The energy of the photon is:

`E=hf\\\\E=6.63* 10^(-34)* 2.8* 10^(15)\\\\E=1.85* 10^(-18)\ J`

Energy utilized by the photon is given by :

`hf=W+(1)/(2)mv^2\\\\hf=6.6+3.5\\\\hf=10.1\ eV\\\\hf=10.1* 1.6* 10^(-19)\ J\\\\hf=1.61* 10^(-18)\ J`

Used energy is given by the difference of energy given and the energy utilized. So,

`E=(18.55-16.1)* 10^(-19)\\E=2.45* 10^(-19)\ J\\\\E=(2.45* 10^(-19))/(1.6* 10^(-19))\\\\E=1.53\ eV`

So, energy used to reach the surface is 1.53 eV.