Question

The work function for the following metals is given: Na: 2.75 eV; K: 2.30 eV; Mo: 4.17 eV; Ni: 5.15 eV. Which of these metals will not give photoelectric emission for a radiation of wavelength 3300 from a He-Cd laser placed 1 m away from the photocell? What happens if the laser is brought nearer and placed 50 cm away ?

Answer 1

Here, the wavelength of radiation is $\lambda =3300A^o=3300\times {10}^{-10}$.

Therefore, the energy of a photon of the incident light,

$E=\frac{hc}{\lambda }=\frac{6.62\times {10}^{-34}\times 3\times {10}^8}{3300\times {10}^{-10}}$

$\Rightarrow 6.018\times {10}^{-19}J$

$\Rightarrow \frac{6.018\times {10}^{-19}}{1.6\times {10}^{-19}}=3.76eV$.

Since the frequency is directly proportional to the energy of the radiation, It is found that the given incident frequency is greater than ${\nu }_0\left(Na\right)$, and ${\nu }_0\left(K\right)$; but less than ${\nu }_0\left(Mo\right)$, and ${\nu }_0\left(Ni\right)$. Therefore, $Mo$ and $Ni$ will not give photoelectric emission. If the laser is brought closer, intensity of radiation increases, but this does not affect the result regarding $Mo$ and $Ni$. However, photoelectric current from Na and K will increase in proportion to intensity.