Question

If 165.7 MJ/mol photon strike atoms of an unknown element and eject electrons with kinetic energies of 25.4 MJ/mol, what is the workfunction of the element? 

What is the frequency of light attributed to the n=3 line in the Balmer transition (n=2) of hydrogen emission

Answer 1

Solution

Using Einstein equation

$E=\phi+KE$

So work function

$\phi=E-KE=165.7-25.4\\=140.3MJ/mol$

Wavelength is given by

$\frac{1}{\lambda}=R(\frac{1}{n^2_f}-\frac{1}{n^2_i})$

Here

$n_i=3\\n_f=2$ putting all value in above equation

Wavelength is found

$\lambda=656.88nm$

So frequency is found

$\nu=\frac{c}{\lambda}=\frac{3\times10^8}{656.88nm}\\=4567\times 10^{11}Hz$