Question

The threshold frequency for caesium metal is $5.16\times { 10 }^{ 14 }Hz$. Determine its work function in eV.

Answer 1

Hint: Work function can be calculated by multiplying Planck’s constant with the threshold frequency. But the answer you get will be in terms of Joules. So to get the answer in terms of eV, divide the obtained answer by charge of an electron which is $1.6 \times { 10 }^{ -19 }C$.

Formula used:
$W= h\nu$

Complete step-by-step answer:
Given: Threshold frequency ($\nu$)= $5.16 \times { 10 }^{ 14 }Hz$
Relation between work function and threshold frequency is given by,
$W= h\nu$ …(1)
where, W: Work function in Joules(J)
            h: Planck Constant in Joules per second (${J}/{s}$)
            $\nu$: Threshold frequency in Hertz(Hz)
By substituting values in equation.(1) we get,
$W= 6.63 \times { 10 }^{ -34 } \times 5.16 \times { 10 }^{ 14 }$
$W= 34.16 \times { 10 }^{ -20 }J$
Now, for converting from Joules to eV, divide the answer by $1.6\times { 10 }^{ -19 }$
$\Rightarrow W= \dfrac { 34.16 \times { 10 }^{ -20 } }{ 1.6 \times { 10 }^{ -19 } }$
$\Rightarrow W=2.13eV$
Hence, the work function of caesium material is 2.13eV.

Additional Information:
Work Function is defined as the minimum energy which is required to emit an electron from the target material. The threshold frequency is defined as the minimum frequency required by electromagnetic radiation to emit an electron. Work functions vary from metal to metal. Caesium has the lowest work function. While platinum is the metal with the highest known work function which is around 6.35eV.

Note: Make sure you convert the work function from Joules to eV correctly. Sometimes, instead of dividing the obtained answer by the charge of electron students multiply them which will lead to a wrong answer. So do take care of that. Note that every metal has different work function and different threshold frequency.