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

Answer:

The work function of potassium is 1.94 eV.

Explanation:

Given that,

Wavelength $\lambda= 3500\ \AA$

Kinetic energy = 1.6 eV

We need to calculate the energy

Using formula of Energy

$E=\dfrac{hc}{\lambda}$

Put the value into the formula

$E=\dfrac{6.63\times10^{-34}\times3\times10^{8}}{3500\times10^{-10}}$

$E=5.68\times10^{-19}\ J$

$E=5.68\times10^{-19}\times6.24\times10^{18}\ ev$

$E=3.54\ ev$

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.