Question

When radiation of frequency 5.6×10^14 Hz are incident on a material, photoelectron are released from it with zero kinetic energy . Calculate the work function of the material .(plank constant 6.6×10^-34 js)

Answer 1

Answer:

The work function of the material will be $3.696\times10^{-19}$.

Explanation:

Given that,

Frequency $f = 5.6\times10^{14}\ Hz$

Kinetic energy $k= 0$

The work function is defined as:

$h\nu=\phi+E_{k}$

Here,

$\phi$= work function

h= plank constant

$\nu$=frequency

$E_{k}$=kinetic energy

The work function will be

$\phi =h\nu-E_{k}$

$\phi=6.6\times10^{-34}\times5.6\times10^{14}-0$

$\phi=3.696\times10^{-19}$

Hence, The work function of the material will be $3.696\times10^{-19}$.