Question

the work function of a metal surface is 3.4eV. what is the frequency of a photon which ejects an electron from this surface at 220 km/s?​

Answer 1

Answer:

7.9*10^14

Explanation:

einsteins equation

hv=hv0+ke

Answer 2

Given:

The work function of a metal surface is 3.4eV.

To find:

Frequency of a photon which ejects electron from the metal surface at a speed of 220 km/s.

Calculation:

Applying Einstein's Equation of Photoelectric Effect:

$\therefore \: KE = h \nu - w_{0}$

$= > \: \dfrac{1}{2}m {v}^{2} = h \nu - w_{0}$

$= > \: \dfrac{1}{2}(9.1 \times {10}^{ - 31})( {220000}^{2} ) = h \nu - (3.4 \times 1.6 \times {10}^{ - 19} )$

$= > \: \dfrac{1}{2}(9.1 \times {10}^{ - 31}){(22)}^{2} \times {10}^{8} = h \nu - (3.4 \times 1.6 \times {10}^{ - 19} )$

$= > \: \dfrac{1}{2}(9.1 \times {10}^{ - 23}){(22)}^{2} = h \nu - (3.4 \times 1.6 \times {10}^{ - 19} )$

$= > \: \dfrac{1}{2}(9.1 \times {10}^{ - 23})(484) = h \nu - (3.4 \times 1.6 \times {10}^{ - 19} )$

$= > \: 0.22\times {10}^{ - 19} = h \nu - (3.4 \times 1.6 \times {10}^{ - 19} )$

$= > \: h \nu = (5.66\times {10}^{ - 19} )$

$= > \: 6.63 \times {10}^{ - 34} \times \nu = (5.66\times {10}^{ - 19} )$

$= > \: \nu = 0.83 \times {10}^{ - 15} \: m$

So, final answer is:

$\boxed{ \bf{ \: \nu = 0.83 \times {10}^{ - 15} \: m}}$