Question

A metal begins to emit electrons, as measured in an apparatus similar to the one Hertz used, when exposed to light at a wavelength of 342 nm (ultraviolet). What is the work function of this metal?

Answer 1

Answer : The work function of this metal is, $5.81\times 10^{-19}J$

Explanation : Given,

Wavelength of light = $342nm=342\times 10^{-9}m$

Formula used :

$E=h\nu_o=\frac{hc}{\lambda}$

where,

E = work function of metal

h = Planck's constant = $6.626\times 10^{-34}Js$

$\nu_o$ = threshold frequency

$\lambda$ = wavelength of light

c = speed of light = $3\times 10^8m/s$

Now put all the given values in this formula, we get the value of work function of this metal.

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

$E=\frac{(6.626\times 10^{-34}Js)\times (3\times 10^8m/s)}{(342\times 10^{-9}m)}$

$E=5.81\times 10^{-19}J$

Therefore, the work function of this metal is, $5.81\times 10^{-19}J$