Question

If threshold wavelength for ejection of electron from metal is 330nm, then work function for photoelectric emission is

Answer 1

Work function is the minimum energy required to eject an electron from the metal surface.
work function=hv°
Threshold frequency=c/^=3×10^8/330×10^-9=1×10^17/11=8^-1
work function=6.626×10^-34×10^17/11=6×10^-19J

Answer 2

The work function is $\bold{6.02 \times 10^{-19} \ J}$

Given:

Threshold wavelength = 330 nm

Solution:

The minimum energy applied on the metal surface to release electron is called as work function. Since it is an energy, the unit of work function is joule (J).

$\text { Work function }(\varphi)=h v_{0}$

Where,

h = Planck’s constant

$v_{0}$ = Threshold frequency

$v_{0}=\frac{c}{\lambda_{0}(G i v e n)}$

On substituting the given value, we get,

$\Rightarrow v_{0}=\frac{3 \times 10^{8}}{330 \times 10^{-9}}=9.1 \times 10^{14} \ \mathrm{m}$

Thus, the work function of photoelectric emission is,

$\varphi=6.626 \times 10^{-34} \times 9.1 \times 10^{14}$

$\therefore \varphi=6.02 \times 10^{-19} \ \mathrm{Joule}$