Question

A metal was irradiated by light of frequency $3.25\times {10}^{15}{S}^{-1}$. The photoelectron produced had its KE, 2 times the KE of the photoelectron which was produced when the same metal was irradiated with a light of frequency $2.0\times {10}^{15}{S}^{-1}$. The work function is:

[Given: $N_A=6\times {10}^{23}$]

• A. 298.35 KJ/mol
• B. 283.23 KJ/mol
• C. 324.56 KJ/mol
• D. 398.78 KJ/mol

Answer 1

The correct option is A 298.35 KJ/mol

Total energy $=h\nu =$ Kinetic energy + work function
Let E and W represent the kinetic energy and work function.
For the frequency $2.0\times {10}^{15}{S}^{-1}$, the expression becomes $E+W=h(2.0\times {10}^{15}{S}^{-1})$......(1)
For the frequency $3.25\times {10}^{15}{S}^{-1}$, the expression becomes $2E+W=h(3.25\times {10}^{15}{S}^{-1})$......(2)
Multiply equation (1) with (2).
$2E+2W=2H(2.0\times {10}^{15}{S}^{-1})$......(3)
Subtract equation (2) from equation (3).
$2W-W=2h(2.0\times {10}^{15}{S}^{-1})-h(3.25\times {10}^{15}{S}^{-1})=6.626\times {10}^{-34}\times 0.75\times {10}^{1}5=4.969\times {10}^{-19}J=4.969\times {10}^{-22}kJ$
This is the work function for one atom.
The work function for one mole of atoms is $6.023\times {10}^{23}\times 4.693\times {10}^{-22}=298.35kJ/mol$.