Question

A particular metal has a work function of 1.05 eV. A light is shined onto this metal with a corresponding wavelength of 324 nm. What is the maximum velocity of the photoelectrons produced? (Hint: 1eV = 1.6022 x 10-19 J, mass of an electron = 9.11 x 10-31 kg)

Answer 1

Answer:

Velocity of electron will be $v=0.986\times 10^6m/sec$              

Explanation:

We have given work function of metal $\Phi =1.05eV=1.05\times 1.6\times 10^{-19}J=1.68\times 10^{-19}J$

Wavelength of the light $\lambda =324nm=324\times 10^{-9}m$

So energy is given by $E=\frac{hc}{\lambda }$, here h is plank's constant and c is speed of light

So $E=\frac{6.6\times 10^{-34}\times3\times 10^8}{324\times 10^{-9} }=6.11\times 10^{-19}j$

For a metal we know that $E=\Phi +KE_{MAX}$

So $KE_{MAX}=E-\Phi =6.11\times 10^{-19}-1.68\times 10^{-19}=4.43\times 10^{-19}$

Now kinetic energy is given by $KE=\frac{1}{2}mv^2$

$4.43\times 10^{-19}=\frac{1}{2}\times 9.11\times 10^{-31}v^2$

$v=0.986\times 10^6m/sec$

So velocity of electron will be $v=0.986\times 10^6m/sec$