Question

Ultraviolet light of wavelength 3500 Angstrom falls on a potassium surface. The maximum energy of the emitted photoelectrons is 1.6 eV. Find the work function of potassium.

Answer 1

The work function of potassium is 1.94 eV.

Step-by-step explanation:

Given that,

Wavelength
`\lambda= 3500\ \AA`

Kinetic energy = 1.6 eV

We need to calculate the energy

Using formula of Energy

`E=(hc)/(\lambda)`

Put the value into the formula

`E=(6.63*10^(-34)*3*10^(8))/(3500*10^(-10))`

`E=5.68*10^(-19)\ J`

`E=5.68*10^(-19)*6.24*10^(18)\ ev`

`E=3.54\ ev`

We need to calculate the work function of potassium

Using formula of work function

`E_(x)=\phi+k_(max)`

Put the value into the formula

`3.54=\phi+1.6`

`\phi=3.54-1.6`

`\phi=1.94\ ev`

Hence, The work function of potassium is 1.94 eV.