Question

The minium energy required for the emisson of a metal elctrons is $133.24\times {10}^{-16}J$. Calculate the cirtical frequency and the correspoding wavelength of he photon (thesehold wavelength) required to eject the electron?
A light source of wavelength $\lambda$ illuinates a metal and ejects photo electrons with $(KE{)}_{max}=1eV$.
Another light source of wavelength $\frac{\lambda }{3}$, ejects photoelectrons from same metal with $\left(KE\right)=SeV$. Find the value of work function $\left(eV\right)$ of metal.

Answer 1

For the first question

eV(Ionisation energy)$=133.24\times {10}^{-16}J$

We use the formula

$eV=hf$ (Eq 1)

where $eV$ is the ionisation energy

$h$ is the Planck's constant$=6.62\times {10}^{-34}$

$f$ is the frequency $f=\frac{C}{\lambda }$

Where C is the speed of light $C=3\times {10}^{8}mse{c}^{-1}$

$\lambda$ is the wavelength

Putting the value of f in equation 1

$eV=h\frac{C}{\lambda }$

$133.24\times {10}^{-16}=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{\lambda }$

$\lambda =1.49\times {10}^{-11}m$

For the second question

$K=hv-h{v}_0$ (eq 1)

where h is the Planck's constant and $v_0$ is the threshold frequency and $h{v}_0$ is the work function and $K$ is the kinetic energy

$v=\frac{C}{\lambda }$

Putting the values in euation 1

$1=\frac{hC}{\lambda }-h{v}_0$ (eq 2)

$4=\frac{3hC}{\lambda }-h{v}_0$ (eq 3)

eq $2$ is multiplied by $-3$ then adding eq $3$ from eq$2$

$-3=\frac{-3hC}{\lambda }+3h{v}_0$

$4=\frac{3hC}{\lambda }-h{v}_0$

$4-3=(3-1)h{v}_0$

$h{v}_0=0.5e{V}^0$