Question

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (νo) and work function (wo) of the metal.

Answer 1

kinetic energy =hc/wavelength - work function
since v=0 so kinetic energy its zero.
so hc/lambda= work function
6.6x10^-34 x 3.0x10^8 / 6800x10^-10 = work function
on solving,
work function= 3x10^-19
Also work function=hx(vo)
so frequency = work function/h
. =3x10^-19 / 6.6x10^-34
. =4.5x10^14

Answer 2

Answer:

The threshold frequency (νo)  is $4.4117\times 10^{14} s^{-1}$ and work function (wo) $2.9117\times 10^{-19} J$ of the metal.

Explanation:

Wavelength of the radiation = 6800 Å = $6800\times 10^{-10} m$

Threshold frequency of the metal = $v_o$

$v_o=\frac{3\times 10^8 m/s}{6800\times 10^{-10} m}=4.4117\times 10^{14} s^{-1}$

Work function = $\phi =hv_o$

$\phi=6.6\times 10^{-34} js\times 4.4117\times 10^{14} s^{-1}=2.9117\times 10^{-19} J$

The threshold frequency (νo)  is $4.4117\times 10^{14} s^{-1}$ and work function (wo) $2.9117\times 10^{-19} J$ of the metal.