Question
The threshold frequency for photoelectric effect on sodium corresponds to a wavelength of $$5000\,\mathop {\text{A}}\limits^ \circ .$$ Its work function is
A.
$$4 \times {10^{ - 19}}J$$
B.
$$1\,J$$
C.
$$2 \times {10^{ - 19}}J$$
D.
$$3 \times {10^{ - 19}}J$$
Answer :
$$4 \times {10^{ - 19}}J$$
Solution :
When a photon of light of frequency $$\nu $$ incident on a photosensitive metal surface, the energy of the photon $$\left( {h\nu } \right)$$ is spent in two ways. A part of energy of photon is used in liberating the electron from the metal surface which is equal to the work function $${W_0}$$ of the metal.
$${W_0} = h{\nu _0}$$
(where, $${{\nu _0}}$$ is threshold frequency)
$$\eqalign{
& {\text{or}}\,\,{W_0} = \frac{{hc}}{{{\lambda _0}}} \cr
& {\text{Here,}}\,\,{\lambda _0} = 5000\,\mathop {\text{A}}\limits^ \circ \cr
& = 5000 \times {10^{ - 10}}m \cr
& \therefore {W_0} = \frac{{6.63 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{5000 \times {{10}^{ - 10}}}} \cr
& = 4 \times {10^{ - 19}}J \cr} $$
NOTE
Threshold frequency $$\left( {{\nu _0}} \right) \to $$ It is the minimum frequency given to metallic surface so that emission of electrons start.