Question

A metal was irradiated by light of frequency $3.25\times {10}^{15}{s}^{-1}$. The photoelectron produced had its kinetic energy, 2 times the kinetic energy of the photoelectrons which was produced when the same metal was irradiated with a light of frequency $2\times {10}^{15}{s}^{-1}$. What is its work function
[Given: $N_A=6\times {10}^{23},h=6.6\times {10}^{-34}Js$]

• A. $297\text{kJ/mol}$
• B. $900\text{kJ/mol}$
• C. $385\text{kJ/mol}$
• D. none of these

Answer 1

The correct option is A $297\text{kJ/mol}$

So the two cases that we have, are:

Using photoelectric equation:
$E=\varphi +KE$ or
$h\nu =\varphi +KE$
where $E=$ energy of incident radiation
$\varphi =$ work function of metal
$KE=$ kinetic energy of ejected electrons

in first case we have :
$h\times 3.25\times {10}^{15}=\varphi +K{E}_1\dots \dots \left(1\right)$
and $K{E}_1=2K{E}_2$

in second case we have :
$h\times 2\times {10}^{15}=\varphi +K{E}_2\dots \dots \left(2\right)$
$K{E}_2=h\times 2\times {10}^{15}-\varphi$

putting value of $K.E_2$ in equation (1) :
$h\times 3.25\times {10}^{15}=\varphi +2(h\times 2\times {10}^{15}-\varphi )$

$h\times 3.25\times {10}^{15}=\varphi +h\times 4\times {10}^{15}-2\varphi$
$\varphi =h\times 0.75\times {10}^{15}$
$\varphi =6.6\times {10}^{-34}\times 0.75\times {10}^{15}\text{J/atom}$
$\varphi =4.95\times {10}^{-19}\text{J/atom}$
$\varphi =4.95\times {10}^{-19}\times \frac{6\times {10}^{23}}{{10}^3}\text{kJ/mol}$
$=297\text{kJ/mol}$